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Southern  Branch 
of  the 

University  of  California 

Los  Angeles 

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INTRODUCTORY 
MATHEMATICAL  ANALYSIS 


BY 


W.  PAUL  WEBBER,  PH.D. 

Assistant  Professor  of  Mathematics  in  the  University  of  Pittsburgh 


AND 


LOUIS  CLARK  PLANT,  M.Sc. 

Professor  of  Mathematics  in  Michigan  Agricultural  College 


FIRST   EDITION 
FIRST    THOUSAND 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC. 

LONDON:   CHAPMAN   &  HALL,  LIMITED 
1919 


COPYBIGHT,    1919, 
BY 

W.  PAUL  WEBBER 

AND 

LOUIS  C.  PLANT 


Stanbope  jpress 

.    H.GILSON   COMPANY 
BOSTON,  U.S.A. 


Asy 

Wsg 


PREFACE 


The  present  course  is  the  result  of  several  years  of  study  and 
trial  in  the  classroom  in  an  effort  to  make  an  introduction  to 
college  mathematics  more  effective,  rational  and  better  suited 
to  its  place  in  a  scheme  of  education  under  modern  conditions 
of  life.  A  broader  field  has  been  attempted  than  is  customary 
in  books  of  its  class.  This  is  made  possible  by  certain  principles 
which  controlled  the  construction  of  the  text. 

One  principle  on  which  the  course  is  built  is  correlation  by 
topics.  For  example,  all  methods  of  calculation  have  been 
associated  in  one  chapter  and  early  in  the  course  in  order  to  be- 
available  for  use  in  the  sequel. 

The  function  idea  has  also  been  emphasized  and  used  as  a 
means  of  correlation. 

Brevity  and  directness  of  treatment  have  contributed  to 
reduce  the  size  of  the  book. 

An  effort  has  been  made  to  keep  in  view  of  the  student  the 
steps  in  the  development  of  the  subject  and  to  point  out  useful 
contacts  of  mathematics  with  affairs. 

The  first  two  chapters  are  intended  to  be  used  for  review  and 
reference  at  the  discretion  of  the  instructor. 

Graphic  representation  and  its  uses  have  been  given  consider- 
able attention.  The  simple  cases  of  determining  empirical 
formulae  give  a  very  valuable  drill  in  the  solution  of  simul- 
taneous equations  and  a  foundation  for  later  work  in  the 
laboratory. 

The  treatment  of  the  trigonometric  functions  is  brief,  direct 
and  in  some  respects  more  advanced  in  style  than  is  customary 
in  current  texts  in  trigonometry  which  are  constructed  mostly 
from  the  secondary  school  standpoint.  Large  use  of  the  func- 


iv  PREFACE 

tions  is  made  in  a  variety  of  applications  in  immediately 
following  chapters. 

More  than  usual  attention  is  given  to  vectors.  The  value 
and  convenience  of  vector  methods  in  science  and  engineering 
seem  to  justify  this  emphasis.  The  part  dealing  with  vector 
products  and  the  problems  depending  on  it  may,  however,  be 
omitted  without  inconvenience  in  later  chapters. 

The  chapter  on  series  may  seem  a  little  heavy  for  freshmen 
but  it  comes  in  the  second  half  of  the  course  and  is  directly 
applied  to  functions  within  the  experience  of  the  student  in  the 
preceding  text. 

What  is  given  on  differential  and  integral  calculus  is  intended 
as  an  introduction  for  those  who  are  to  take  the  regular  and 
fuller  course  in  calculus.  For  those  who  are  not  to  continue 
their  mathematics  it  will  furnish  an  introduction  to  the  methods 
of  calculus  and  some  important  definite  applications.  The 
integral  has  first  been  regarded  as  the  inverse  of  the  derivative 
and  nothing  is  said  about  the  differential.  This  seems  natural 
and  in  accord  with  the  idea  of  the  solution  of  differential  equa- 
tions under  many  actual  conditions  where  a  function  is  sought 
whose  derivative  is  given.  Following,  the  integral  is  regarded 
as  a  summation  of  elements  and  some  further  applications  are 
introduced.  In  the  list  of  integrals  for  reference  both  the  in- 
verse and  the  differential  forms  are  given. 

In  general  no  effort  at  rigor  beyond  reasonable  conviction 
has  been  attempted.  Proofs  have  been  given  for  some  theorems 
that  many  teachers  may  prefer  to  regard  as  assumptions. 
These  proofs  may,  therefore,  be  omitted  at  the  discretion  of 
the  teacher.  A  number  of  what  appear  as  theorems  in  some 
texts  are  here  given  as  exercises.  For  this  reason  it  is  recom- 
mended that  each  student  be  held  for  practically  all  the  exter- 
cises  appearing  regularly  through  the  text.  Selections  may  be 
made  at  the  instructor's  discretion  from  the  exercises  at  the  end 
of  each  chapter. 

The  teacher  will  find  an  opportunity  for  originality  in  develop- 
ing the  text  and  at  times  a  necessity  for  more  details. 


PREFACE  V 

The  entire  course  has  been  given  several  times  to  classes 
meeting  daily  during  a  period  equivalent  to  four  terms  of  three 
months  each.  The  work  can  be  covered  in  less  time. 

The  authors  take  this  opportunity  to  acknowledge  their 
obligation  to  Dean  H.  B.  Meller,  of  the  University  of  Pittsburgh, 
for  affording  the  opportunity  to  have  the  course  tried  out  in 
the  classroom  and  for  a  number  of  problems. 

If  this  book  shall  contribute  toward  making  more  satisfactory 
and  more  economical  the  modicum  of  college  mathematics  the 
authors  will  feel  well  repaid  for  the  considerable  labor  of  its 
composition. 

W.  P.  W. 
L.  C.  P. 


TABLE  OF  CONTENTS 

(Numbers  refer  to  sections) 

CHAPTER  I 
REVIEW  OF  ALGEBRA 

PAGE 

1.  Signs  of  aggregation 1 

2.  Index  laws 2 

3.  Important  cases  of  factoring 3 

4.  Lowest  common  multiple 4 

5.  Highest  common  factor 5 

6.  Fractions 7 

7.  Equations;  root  of  an  equation 8 

8.  Simultaneous  equations  and  elimination 8 

9.  Formulas  relating  to  radicals 11 

10.  Quadratic  equations 12 

11.  Inequalities 14 

12.  Binomial  formula 15 

* 

CHAPTER  II 
GEOMETRICAL  THEOREMS 

13.  Geometrical  theorems  and  formulas 16 

CHAPTER  III 
METHODS   OF  CALCULATION 

14.  Need  of  numbers  and  calculations 20 

15.  Important  discoveries  relating  to  numbers 15 

16.  Mechanical  devices  for  calculating 21 

17.  Graphic  or  geometric  representation  of  numbers;  scale 22 

18.  Arithmetical  operations  by  geometric  methods 22 

19.  Idea  of  logarithms 24 

20.  Definition  of  logarithm 25 

21.  Rules  for  calculating  with  logarithms ' 25 

22.  Characteristic  and  mantissa  of  a  logarithm 26 


viii  TABLE  OF  CONTENTS 

PAOE 

23.  Use  of  tables  of  logarithms  in  calculating 28 

24.  Exponential  equations  solved  by  use  of  logarithms 28 

25.  Logarithmic  scale;  slide  rule 30 

26.  Rules  for  calculating  with  Mannheim  slide  rule 31 

26a.  Double  interpolation 31 

CHAPTER  IV 
GRAPHIC  REPRESENTATION 

27.  Graphic  representation  of  statistical  data 36 

28.  Axes  and  coordinates  defined 37 

CHAPTER  V 
RATIO   PROPORTION  AND  VARIATION 

29.  Proportion 45 

30.  Ratio;  measurement 45 

31.  Formulas  of  proportion 46 

32.  Variation;  direct,  inverse  and  joint 46 

CHAPTER  VI 

RECTANGULAR  COORDINATE  SYSTEM;    GRAPHIC  REPRESEN- 
TATION OF  EQUATIONS 

33.  Axes  and  coordinates;  quadrants 52 

34.  Graphs  of  functions  and  equations 53 

35.  Emperical  equations  or  formulas 57 

CHAPTER  VII 
NUMBERS,  VARIABLES,  FUNCTIONS  AND   LIMITS 

36.  Classes  of  numbers 62 

37.  Variable;  function;  sequence;  limit 63 

38.  Functional  notation 65 

39.  Problem  of  mathematics 66 

40.  Increment  of  a  variable 66 

41.  Special  forms  and  limits 66 

42.  Proofs  of  theorems 67 

43.  Limiting  values  of  expressions  for  certain  values  of  the  variable  68 

44.  Idea  of  function  developed  from  number  pairs  and  curves 69 


TABLE  OF  CONTENTS  ix 


CHAPTER  VIII 

CIRCULAR   (TRIGONOMETRIC)   FUNCTIONS  AND   THEIR 
APPLICATIONS 

PAGE 

45.  Problem 71 

46.  Angle  defined 72 

47.  Trigonometric  ratios  defined 73 

48.  Fundamental  formulas 75 

49.  Functions  at  the  quadrant  limits 78 

50.  Functions  of  negative  angles 79 

51.  Complement  relations 80 

52.  Reduction  to  functions  of  angles  less  than  90° 81 

53.  Addition  theorems 85 

54.  Trigonometric  equations;  inverse  functions 89 

55.  Formulas  relating  to  right  triangles 91 

56.  Directions  for  solving  problems 91 

57.  Sine  law 95 

58.  Cosine  law 95 

59.  Example  of  the  use  of  the  sine  law 96 

60.  Example  of  the  use  of  the  cosine  law 97 

61.  Conversion  formulas 98 

62.  Tangent  law 99 

63.  Ambiguous  cases 100 

64.  Double  and  half  angles 101 

65.  Angle  of  a  triangle  in  terms  of  the  sides 102 

66.  Radius  of  inscribed  circle 103 

67.  Radius  of  circumscribed  circle 103 

68.  Radian  measure  of  angles 104 

68a.  Mil  as  a  unit  of  angular  measure 106 

69.  Explanatory  definitions  relating  to  field  work 106 

70.  Graphs  of  trigonometric  functions 108 

71.  Graphs  of  inverse  functions 110 

72.  Equations  involving  trigonometric  functions  as  unknowns 112 

72a.  Special  interpolations  and  tables 113 


CHAPTER  IX 
POLAR  COORDINATES;    COMPLEX  NUMBERS;    VECTORS 

73.  Polar  coordinates 120 

74.  Powers  of  V  —  1  =  i 122 

75.  Geometrical  representation  of  complex  numbers 122 


X  TABLE  OF  CONTENTS 

PAGE 

76.  Arithmetical  operations  with  complex  numbers 123 

77.  Multiplication  and  division  in  polar  form 124 

78.  Vector  quantities;  vectors 125 

79.  Rectangular  and  polar  notations  of  vectors 125 

80.  Addition  and  subtraction  of  vectors 126 

81.  Multiplication  of  vectors 128 

82.  Components  of  vectors  on  axes 130 

83.  Equilibrium  of  particles  and  rigid  bodies 132 

CHAPTER  X 
EQUATIONS 

84.  Integral  equation;  factor  theorem 139 

85.  Fundamental  theorem  of  algebra 139 

86.  Identity  theorem 140 

87.  Remainder  theorem 141 

88.  Zero  and  infinite  roots 141 

89.  Synthetic  division 142 

90.  Theorem  on  root  of  an  equation 143 

91.  Solution  of  numerical  equations 144 

92.  Quadratic  equations 149 

93.  Equations  in  quadratic  form 150 

CHAPTER  XI 
LINEAR  FUNCTIONS  AND   THE  STRAIGHT  LINE 

94.  General  linear  function 152 

95.  Theorem  on  the  linear  equation  and  the  straight  line 152 

96.  Families  of  straight  lines 153 

97.  Converse  theorem 154 

98.  Normal  form  of  the  equation  of  the  straight  line 156 

99.  Different  forms  of  equation  of  the  straight  line 157 

100.  Distance  between  two  points 158 

101.  Division  of  a  line  segment 158 

102.  Angle  between  two  lines 160 

CHAPTER  XII 

EQUATIONS  OF  THE  SECOND  AND  HIGHER  DEGREES  AND 
THEIR  GRAPHS 

103.  Explicit  and  implicit  functions  defined 1 63 

104.  Explicit  quadratic  function;  discussion  of  a  curve 163 


TABLE  OF  CONTENTS  xi 

PAGE 

105.  Implicit  quadratic  function 165 

106.  Reference  to  conic  sections 166 

107.  Functions  of  the  third  degree 167 

108.  Rational  fractional  function 167 

109.  Irrational  function 168 

110.  Simultaneous  equations  of  the  second  and  higher  degrees 169 

111.  Equivalence  of  equations 169 

112.  Some  systems  of  quadratic  equations 171 

CHAPTER  XIII 
TRANSFORMATION   OF  COORDINATES 

113.  Linear  transformation;  rotating  the  axes;  moving  the  origin. . .  174 

CHAPTER  XIV 
CONIC  SECTIONS 

114.  Conic  sections  defined 177 

115.  Parabola 177 

116.  Ellipse 179 

117.  Hyperbola 181 

118.  Diameters 183 

119.  Eccentric  angle 185 

120.  General  equation  of  the  second  degree  in  two  variables 187 

121.  Confocal  conies 188 

122.  Centers  of  conies 189 

CHAPTER  XV 

THEOREMS   ON  LIMITS;    DERIVATIVES  AND  THEIR 
APPLICATIONS 

123.  Limit  of  sum  of  infinitesimals 193 

124.  Limit  of  sum  of  variables 193 

125.  Limit  of  product  of  variables ,. 194 

126.  Limit  of  quotient  of  two  variables 194 

127.  Definition  and  formation  of  derivative 194 

128.  Reference  to  special  rules 197 

129.  Derivative  of  constant 197 

130.  Derivative  of  variable  with  respect  to  itself 197 

131.  Derivative  of  positive  integral  power  of  function 198 

132.  Extension  to  positive  fractional  power 198 


xii  TABLE  OF  CONTENTS 

PAGE 

133.  Extension  to  negative  power 198 

134.  Extension  to  irrational  power 199 

135.  Extension  to  imaginary  power 199 

136.  Derivative  of  product  of  functions 199 

137.  Derivative  of  quotient 200 

138.  Derivative  from  an  implicit  function 200 

139.  Slope  of  a  curve  at  a  point;  equation  of  tangent  line 202 

140.  Maximum  and  minimum  values  of  functions 205 

141.  Use  of  derivative  to  determine  maximum  and  minimum  values  205 

142.  Increasing  and  decreasing  functions;   conditions  for  maximum 

and  minimum  values 207 

143.  Use  of  second  derivative  to  determine  maximum  and  minimum 

values .' 208 

144.  Use  of  derivative  to  define  motion 214 

145.  Use  of  derivative  to  discover  equal  roots  of  equations 217 

146.  Force  the  derivative  of  momentum  with  respect  to  time 218 

CHAPTER  XVI 
SERIES;    TRANSCENDENTAL  FUNCTIONS 

147.  Sequences 219 

148.  Series  defined;  convergence  defined 219 

149.  Arithmetic  series 220 

149a.  Geometric  series 222 

150.  Special  case  of  geometric  series 224 

151 .  Harmonic  series 225 

152.  Convergence  of  series . . '. 226 

153.  Comparison  test  of  convergence 228 

154.  Standard  series  for  comparison 228 

155.  Ratio  test  of  convergence 230 

156.  Series  with  complex  terms 231 

157.  Expansion  of  functions  in  power  series 232 

158.  Functions  expanded   about   the  origin;    functions  expanded 

about  any  point;  formulas 232 

159.  Binomial  expansion,  any  exponent. 234 

160.  Exponential  function  and  series 236 

161.  Theorem  on  logarithms 236 

162.  Derivative  of  exponential  function;   derivative  of  logarithmic 

function 237 

163.  Use  of  logarithm  in  calculating  derivatives 238 

164.  Logarithmic  series;  calculation  of  logarithms 239 

165.  Exponential  (Euler's)  values  of  sine  and  cosine 240 


TABLE  OF  CONTENTS  xiii 

PAGE 

166.  Derivatives  of  sine  and  cosine 241 

167.  Expansion  of  sine  and  cosine  in  series 242 

167a.  Logarithmic  graphs 244 

167b.  Empirical  formulas  derived  from  logarithmic  graph 246 

167c.  Empirical  formulas  derived  from  semilogarithmic  graph 247 


CHAPTER  XVII 
INTEGRATION 

168.  Integral  defined 250 

169.  Directions  for  solving  problems 253 

170.  Definite  integral 254 

171.  Area  under  a  curve 256 

172.  Volume  of  solid  of  revolution 257 

173.  Average  value  of  a  function  over  an  interval  of  the  variable.  . .  259 

174.  Work  of  a  variable  force 261 

175.  Integral  regarded  as  a  sum  of  infinitesimal  elements 263 

176.  Centroids  determined  by  integration 264 

177.  Integration  by  parts 267 

178.  Length  of  an  arc  of  a  curve 268 

179.  Simplifying  integrand  by  transformation  of  variable 269 

Additional  formulas 271 

Supplementary  exercises 271 

Table  of  integrals 276 

Use  of  tables 280 

Calculating  tables 284 


INTRODUCTORY 
MATHEMATICAL  ANALYSIS 

CHAPTER  I 
REVIEW  OF  ALGEBRA 

1.  Signs  of  aggregation  are  used  to  indicate  that  an  opera- 
tion is  to  extend  to  each  term  of  a  group  of  terms  or  to  separate 
or  to  specify  torms  or  factors  in  an  algebraic  expression.  Thus: 

1.  3  a  (6  +  c  +  d)  shows  that  each  term  inside  the  "  (      ) " 
is  to  be  multiplied  by  the  factor,  3  a,  which  precedes  the  "  (      )  ", 
or  3  a  (b  +  c  +  d)  is  equivalent  to  3  ab  +  3  ac  +  3  ad. 

2.  V6  a  +  3  b  -  24  shows  by  the  " "  (vinculum)  that 

the  "V"  (square  root)  is  to  be  taken  of  the  expression  under 

the  " "  considered  as  a  whole,  and  not  the  square  root  of 

each  term  separately.  

3.  a  —  (2  c  —  4  &  +  d)  or  a  —  2  c  —  4  6  +  d  shows  that  the 

terms  in  '•'  (      ) "  or  under  " "  are  all  to  be  subtracted  from 

a.    Since  we  may  either  subtract  these  singly  or  first  combine 
them  and  subtract  the  combined  result,  we  may  express  the 
operation  as, 


=  a-2c-  (-46)  -d=a-2c  +  4&-d. 

4.  (3  -f-  6  a  —  4  6)  (a  —  5  c  +  d)  shows  that  the  combined 
value  of  the  terms  in  each  "  (      ) "  is  to  be  multiplied  by  the 
other.     This  is  accomplished  by  multiplying  every  term  in  one 
"  (      ) "  by  each  term  in  the  other  "  (      )  "  and  reducing  the 
result  to  the  simplest  form. 

5.  a  -\-  b  —  c  +  d  +  Vc  —  e  may,  by  using  a  sign  of  aggre- 
gation, say  a  " (      ),"  be  written  as  a  +  &  +  d  —  (c  +  e 

1 


2  REVIEW  OF  ALGEBRA 

Remark.  —  When  a  sign  of  aggregation  is  preceded  by  a 

sign,    the    sign    of    aggregation    may    be    removed 
mill  us 


the  si£ns  of  the  terms  within  the  si^  of 
aggregation.  We  must  not  confuse  signs  of  aggregation  with 
signs  of  operation.  Thus,  in  the  expression  —  Va  —  6,  the 
"V"  is  not  a  sign  of  aggregation,  but  shows  that  the  square 
root  is  to  be  taken  of  the  entire  expression  under  the  "— 
considered  as  a  whole.  The  minus  sign  before  the  "V"  is  not 
to  be  regarded  as  preceding  the  sign  of  aggregation  and  there- 
fore the  "  -  "  in  the  expression  cannot  be  removed  by  the 
usual  rule.  The  operation  of  square  root  must  first  be  per- 
formed. Similarly,  other  cases  are  to  be  handled. 


I2x  -  (9x  +  7x)  =  ?;  ofc  +  3  -  a&  =  ? 


2.  xy  -  Vxz-2x+l  =  ?  ;  o%2  -  x2y2  -  1  =  ? 

3.  a  —  (a  +  &  —  (c  —  d-\-  e  —  a)  +  c)  —  6—  c  +  d  —  e  =  ! 

4.  (ab  ~(cd+  1))  (06  -  (cd  -  1))  =  ? 

5.  (2z2+(3z-l)  (4  3  +  5))  (5z2-(4s  +  3)  (x  -  2))  =? 

6.  x—  {-l2y-[2x+(-4:y-  (-7x-5y)  -Qx-9y 


7.  (a2  -  (62  +  c2))  (a2  -  (62  -  c2))  =  ? 

8.  20  a  (10  a  -  (6  a  -  c  -  '(5  a  -£>)).+  c)  =  ? 

9.  (V9-6z2  +  z4  -  V25-10a  +  a2)  (3  -  \/25)  = 


2.  The  index  laws  may  be  stated  in  the  form  of  equations  as 
follows: 

1.  am  •  dn  =  am+n. 

2.  am  -i-  an  =  am~n. 

3.  am  -i-  am  =  1,  also  am  -5-  am  =  aw-m  =  a°.  ' 

.'.  a°  =  1  for  any  value  of  a. 

4.  1  -T-  am  =  a°  -f-  am  =  a°~m  =  o-»,  but  1  -i-  am  =  I/a*. 

/.  a"*  =  l/am. 

5.  (om)n  =  amn. 
60   (06)  n  =  an6n. 


CASES  OF  FACTORING  3 

Perform  the  indicated  operations  with  the  expressions  just  as 
they  are  written  and  change  the  result  so  that  all  exponents 
shall  be  positive. 

1 .  (a2ar !  +  3  a3x~2)  (4  a~l  -  5  or1  +  6  cur2)  =  ? 

2.  (a°  -  a-1)  (a°  -  cr2)  (a°  -  a-3)  =  ? 

3.  (x*  -  ?/*)  (**  +  x  V  +  r)  =  ? 

4.  (18  y~*  +  23  +  xr*y  +  6arV)  4-  (3  xV'+z*  -  2  x~*y)  =  ? 

5.  (x-27/)6  =  ? 

6.  (3  alb*  +  4 a&3  -  a? 6)  (6 o*H  -  8 a~*H  -  2 a~*)  =  ? 

7.  (8x2-2x-3)/(12x2-25x-12)  =? 

When  x  =  f ,  (a)  by  direct  substitution,  (6)  by  carrying  out 
the  division  and  substituting  in  the  quotient. 

8.  (XP  -  3  x?-1  +  4  x*-2  -  6  x"-3  +  5  a;"-4)  (2  z3  -  z2  +  x)  =  ? 

9.  (x5  -  i/6)  ^-  (x  -  y)  =  ? 

10.  (6  xm~n+z  —  xm~n+1  —  22  xm~n  —  19  x"1"""1  —  4  zm~n~2) 
-^  (3  x3-"  +  4  x2-"  +  or")  =  ? 

3.  Some  important  cases  of  factoring  depend  on  the  fol- 
lowing: 

1.  (a  +  &)  0  +  &)  =  (a  +  6)2  =  a2  +  2  ab  +  62. 

2.  (a  -  6)  (a  -  b)  =  (a  -  6)2  =  a2  -  2a6  +  62. 

3.  (a  +  6)  (a  -  6)  =  a2  -  62. 

4.  (a  +  x)  (a  +  #)  =  a2  +  (x  +  y)  a  +  xy. 
5=  (a  +  b)  (a2  -  ab  +  &2)  =  a3  +  63. 

6.  (a  -  6)  (a2  +  ab  +  62)  =  a3  -  b3. 

7.  (a  ±  6)3  =  a3  ±  3  a26  +  3  a&2  ±  63. 

8.  (a  +  &)n  =  a"  +  wa""1 6  +  (n  (n  -  1)/1  •  2)  an~2 62 
+  (n(n-l)(n-2)/1.2-3)an-3&3  +  -  •  •  +  nab"-1  +  6". 

9.  (an  -  6n)  =  (a  -  6)  (a"-1  +  an~2  &  +  a"-3  b2  +  •  •  •  +  abn~2 

+  &"-1)- 

10.  ax  —  ay  +  bx  —  by  =  a  (x  —  y)  +  b  (x  —  y)  =  (a  +  &)  (x  —  ?/). 

11.  (a±fe±c)2  =  a2  +  &2  +  c2 

Factor  the  following: 

1.  20  db  -  28  ad  -  5  be  +  7  cd. 

2.  49  x6  -  168  x3?/  +  144  y2. 


REVIEW  OF  ALGEBRA 

3.  6  xy  +  16  z2  -  9  y2  -  z2. 

4.  o2-2a6  +  &2-c2. 

5.  a2  -  10  a  -  75. 

6.  a2-2a&  +  62-3a  +  3&  +  2. 

7.  x2  +  2  xy  -  99  i/2. 

8.  36  re2  +12  a;  -35. 

9.  z3-3z2  +  3z-l. 

10.  a3-6a2  +  18a-27. 

11.  8x3-27y3. 

12.  x3-!. 

13.  a9-!. 

14.  a2  —  w2  +  a  +  m. 

15.  (Sz3  -  27)  -  (2z  -  3)  (4z2  +  4z  -  6). 

16.  x3  +  2  a%  +  2  z?/2  +  y3. 

17.  (a2  +  6  a  +  8)2  -  14  (a2  +  6  a  +  8)  -  15. 

18.  x*-x5-  32  ^  +  32. 

19.  18z5&  +  24z3&3  +  8z65. 

20.  27  tf  -  108  x*y  +  144  xy*  -  64  j/». 

21.  21  a2  +  23  06  +  6  62. 

22.  10  a2  -39  a  +  14. 

23.  95-14^-s4. 

24.  x»  +  512. 

25.  5  x  +  25  Z?/ 


4.  The  lowest  common  multiple  of  two  or  more  expressions 
is  that  expression  of  lowest  degree  that  is  exactly  divisible  by 
each  of  the  given  expressions.  Thus: 

1.  a263c  is  the  lowest  common  multiple  of  a2,  a263  and  aWc. 

2.  y?  -f-  63  is  the  lowest  common  multiple  of  x  +  6  and 
x2  -  bx  +  fe2. 

Remark.  —  In  calculating  the  lowest  common  multiple  of 
several  expressions  note  that  it  must  contain  as  factors  every 
different  factor  of  all  the  expressions.  In  case  a  factor  "occurs 
more  than  once  in  any  of  the  expressions  it  is  to  be  taken  to  the 
highest  power  that  it  occurs  in  any  of  the  expressions. 


THE  HIGHEST  COMMON  FACTOR 

Find  the  lowest  common  multiple  of  the  following: 

1.  3z3-13z2  +  23z-21  and  6x3  +  a;2  -44  a;  +  21. 

(Try  3  x  —  7  as  factor.) 

2.  z2  +  x4,  2  xz  -  4  x  and  x2  +  1. 

3.  a2  +  ab  +  ac  +  be  and  a2  +  2  ab  +  62. 

4.  a2  -  15  a  +  50,  a2  +  2  a  -  35  and  a2  -  3  a  -  70. 

5.  z2  +  5  x  +  6,  a?  -  19  x  -  30  and  re3  -  7  z2  +  2  a;  +  40. 

6.  a3  +  6  a2  +  11  a  +  6  and  a4  +  a3  -  4  a2  -  4  a. 

7.  8a2-6a  -9  and  6a3  -7a2-7a  +  6. 

8.  x2  -  7a#  +  12  yz,  xz  -  Qxy  +  8y2  and  z2  - 


6.  The  highest  common  factor  of  two  or  more  expressions  is 
that  expression  of  highest  degree  that  will  be  an  exact  divisor  of 
the  given  expressions.  Thus,  abc  is  the  highest  common  factor 
of  o&2c2,  a?bc,  abc?.  The  highest  common  factor  must  contain 
as  factors  all  the  factors  that  are  common  to  all  the  expressions. 
The  highest  common  factor  is  usually  obtained  by  factoring 
each  of  the  given  expressions  and  then  taking  each  factor  as 
many  times  as  it  is  common  to  all  the  expressions.  The  product 
of  these  factors  is  the  highest  common  factor. 

When  the  expressions  are  not  easily  factored  a  method  due  to 
Euclid  *  is  useful.  Let  A  and  B  be  two  expressions  (or  num- 
bers). Suppose  (1)  A  =  qB  +  Rl}  (2)  B  =  q^  +  Rz, 

(3)  Ri  =  qJh  +  R3,    (4)   R2  =  q3R3. 

From  these  we  have,  since  Rz  =  q^Ra,  Rs  is  the  highest  factor 
of  #2,  Ri,  R3.     Hence  B  =  R3  (q&qa  +  q\+  qz). 

A  =  R3  (qqiqtfs  +  qqi  +  qqs  +  q^z  +  1). 

Therefore,  the  highest  common  factor  of  A,  B  is  R3. 

This  suggests  the  following  rule: 

Divide  A  by  B,  call  the  quotient  q  and  the  remainder  RI. 

Next,  divide  B  by  Ri,  call  the  quotient  q\  and  the  remainder 
Rz. 

Next,  divide  #1  by  Rz,  call  the  quotient  qz  and  so  on,  con- 
tinuing to  divide  the  last  divisor  by  the  last  remainder,  until  a 

*  See  Geometry.    This  method  may  be  omitted  if  desired. 


6  REVIEW  OF  ALGEBRA 

remainder  0  is  obtained.     The  last  divisor  used  is  the  highest 
common  factor  of  A,  B. 

If  it  becomes  evident  during  the  above  process  that  the  final 
division  cannot  give  a  remainder  0,  then  A,  B  have  no  common 
factor  different  from  unity. 

Remark.  —  During  the  process  outlined  above  we  may,  when 
necessary  or  convenient,  divide  or  multiply  any  of  the  expres- 
sions A,  B,  Ri,  R2,  .  .  .  by  any  number  not  a  common  factor 
of  A,  B  without  affecting  the  highest  common  factor.  If  a 
factor  be  used  which  is  common  to  A,  B,  we  must  account  for 
it  in  making  up  the  highest  common  factor  of  A,  B. 

The  following  example  will  illustrate  the  process: 

Find  the  highest  common  factor  of 
Gz3  +  7x2y  -3xy2  and  4x*y  +  8xY  -3xy*  -  9y*. 

First  we  take  x  out  of  the  first  expression  and  y  from  the  second. 
We  then  have 

6xz  +  7xy  -Zy*  and  4s3  +  8x*y  -  3xyz  -  Qy*. 

Since  4  x3  is  not  divisible  by  6  x2  we  multiply  the  second  expres- 
sion by  3.    We  have  then 

6  xz  +  7  xy  -  3  yz)l2  x3  +  24  x*y  -  9  xy*  -  27  j/»(2  x  +  5  y 

-Qxyz 


!Qxzy-Zxy*-27y* 
Multiplying  by  3,  3 


30x2y  +  35  sy2  -  15  y3 


Divide  by  -22  y\  (-44xy*  -  66  y3)  ^  (-22  y2)  =  2x 

2  x  +  3  y)  6  x2  +  7  xy  -  3  y2(3  z  -  y 
6  s2  +  9  xy 


Therefore  2  x  +  3  y  is  the  highest  common  factor  of  the  given 
expressions. 


MANIPULATION  OF  FRACTIONAL  EXPRESSIONS 


Find  the  highest  common  factor  of: 

1.  16,  72,  144. 

2.  1728,  576. 

3.  a2  -  5  ab  +  4  62  and  a3  -  5  o26  +  4  6s. 

4.  2  a2  -  5  a  +  2  and  12  a3  -  8  a2  -  3  a  +  2.4 

5.  z2  -  1,  x3  +  1  and  a;  +  1. 

6.  10  (a:  +  I)3  and  4  (3  +  I)2  (a:  -  1). 

7.  z4  -  3z2  -  28,  x4  -  16  and  z3  +  z2  +  4z  +  4. 

6.   All  the  preceding  processes  may  be  used  in  the  reduction 
and  manipulation  of  fractional  expressions. 

1.  (a2  -  8  06  +  7  62)/(a2  -  3  ab  +  2  62)  to  lowest  terms. 

2.  (x6  —  if)  (x  —  y)/((x*  —  y3)  (x*  —  t/4))  to  lowest  terms. 

3.  (a  +  6)  ((a  +  6)2  -  c2)/(4  aV  -  (a2  -  62  -  c2)2)  to  lowest 
terms. 

4.  (3-x)/(l-3x)  -"(3+a;)/(l+3a;)  -  (l-16z)/(9z2-!), 
combine  and  reduce. 

5.  x/(x  —2y)—  y/(2  y  —x)  —  (x  —  y^/(x2  —  4  7/2),  combine 
and  reduce. 

6.  x/(x  +  2)  -3/(a;  -  4)  +  3/(x  -  6)  -  l/(z  -  8),  combine 
and  reduce. 

7.  (x2  -  yz)/((x  +  y)(x  +  z))  +  (i/2  -  zx)/((y  +  z)  (y  +  x)) 
+  (z2  —  xy)/((z  +  x)  (2  +  t/)),  combine  and  reduce. 

8.  1/(1  -  x)  -JL/0  +s)/l/0  -  z2)  -  1/(1  +  x2),  combine 
and  reduce. 

9.  (2-x-(6x 
combine  and  reduce. 

10.  (27z3 
combine  and  reduce. 

11.  1  _ 
3  +  1*3 


4  +  3-5 


(Begin  at  the  bottom; 
combine  and  reduce.) 


5-7 


12.   a+1 


a+1 


a+1 


a+1 
a 


8  REVIEW  OF  ALGEBRA 

7.  To  solve  an  equation  is  to  find  a  value  of  some  letter  in  the 
equation  which  will,  when  substituted  for  that  letter,  verify  or 
satisfy  the  equation.     The  value  of  a  letter  (unknown  quantity) 
in  an  equation  which  will  satisfy  the  equation  is  called  a  root 
of  the  equation.     Then,  to  solve  an  equation  is  to  find  its  root 
or  roots. 

It  is  assumed  that  the  student  knows  how  to  solve  simple 
equations,  including  equations  containing  fractions.  The 
following  exercises  will  afford  review  and  extension  of  that 
knowledge. 

1.  (6z+  1)/15  -  (2x  -4)/(7z  -  16)  =  (2  x  -  l)/5. 

Hint.  —  Multiply  each  member  by  the  lowest  common 
multiple  of  the  denominators  and  solve.  That  is,  clear  of 
fractions  and  solve  the  equations  for  x. 

2.  l/(x  -  2)  -  l/(x  -  3)  =  l/(x  -  4)  -  l/(x  -  5). 

3.  (x  +  2)/(*  -  3)  +  (x  -  3)/(*  +  4)  -  (x  +  4)/(z  +  2)  =  3. 

4.  (2x-  l)/(2  x  -  3)  -  (x2  -  x)/(x*  +  4)  =  2. 

5.  (x  -  !)/(«  -  '2)  +  (x  -  5)/(z  -  4)  +  (x  -  7)/(z  -  6) 
+  (x  -  7)/(z  -  8)  =  4. 

6.  a/(x  —  a)  —  b/(x  —  6)  =  (a  —  b)/(x  —  c),  where  a,  6,  c  are 
regarded  as  known. 

7.  bx/a  -  (a2  +  62)/a2  =  a2/62  -  x  (a  -  b}/b. 

8.  (x3  +  !)/(*  +  1)  -  (x*  -  l)/(x  -  1)  =  20. 

Note.  —  It  is  better  in  this  case  to  reduce  the  fractions  to 
lowest  terms  before  solving. 

9.  (ax  -  a2)/(x  -  6)  +  (bx  -  tf)/(x  -  a)  =  a  +  6. 

10.  (m  +  ri)/(x  +  m  —  n)  —  2  m/(x  —  m  -f-  ri)  +  (m  —  n)/ 
(x  —  m  —  ri)  =  0. 

8.  A  single  equation  is  sufficient  to  determine  the  value  of 
one  unknown  symbol  in  the  equation.    We  are  to  think  of  an 
equation  containing  one  unknown  as  a  condition  to  be  satisfied 
by  the  unknown.    We  may  impose  one  independent  or  arbi- 
trary assumption  on  one  unknown.     This  assumption  may  be 
put  in  the  form  of  an  equation.     It  is  then  a  problem,  more  or 


A  SINGLE  EQUATION  9 

less  difficult,  to  determine  the  value  of  the  unknown  that  will 
satisfy  the  condition.  That  is,  we  must  find  a  root  of  the 
equation,  or,  as  we  say,  solve  the  equation  for  the  unknown. 

We  may  make  two  independent  assumptions  or  impose  two 
independent  conditions  on  two  unknowns  simultaneously. 
These  conditions  may  be  put  in  the  form  of  equations.  We 
must  then  solve  the  equations  to  determine  the  unknowns.  To 
do  this  we  must  derive  from  the  two  equations  a  single  equation 
with  only  one  unknown.  This  equation  will  when  solved  give 
the  value  of  one  unknown.  By  substituting  this  value  for  that 
unknown  hi  one  of  the  original  equations  there  will  result  an 
equation  with  the  other  unknown  which  may  now  be  deter- 
mined. The  process  of  deriving  a  single  equation  with  one 
unknown  from  two  equations  each  containing  two  unknowns  is 
called  elimination  of  an  unknown  or  symbol. 

When  three  equations  each  containing  three  unknowns  are 
given  for  solution,  we  must  derive  two  equations  each  contain- 
ing the  same  two  unknowns  and  from  these  finally  derive  a 
single  equation  with  only  one  unknown.  The  last  equation 
being  solved  gives  the  value  of  one  unknown.  This  value  when 
substituted  in  one  of  the  two  equations  will  give  an  equation 
from  which  another  unknown  can  be  determined.  Having  the 
values  of  two  unknowns  they  may  be  substituted  in  one  of  the 
original  equations  and  then  from  the  resulting  equation  the  value 
of  the  third  unknown  can  be  determined. 

In  general,  the  process  of  deriving  re  —  1  equations  with 
re  —  1  unknowns  from  n  equations  with  n  unknowns  is  called 
elimination.  The  idea  of  elimination  and  substitution  is  very 
important.  Elimination  is  carried  out  in  elementary  algebra  by 
several  methods.  Experience  only  can  teach  the  pupil  the  best 
method  to  use  in  a  given  case.  The  following  three  methods 
are  the  ones  most  used: 

1.  Combine  two  equations  by  addition  or  subtraction  so  as  to 
cause  one  unknown  to  disappear.  Often  one  or  both  equations 
must  be  multiplied  or  divided  by  some  known  number  in  order 
that  addition  or  subtraction  will  cause  an  unknown  to  disappear. 


10  REVIEW  OF  ALGEBRA 

2.  Solve  one  of  two  equations  for  one  unknown  in  terms  of 
the  other  unknowns  and  substitute  this  value  in  the  other  equa- 
tions.    The  resulting  equations  will  contain  one  less  unknown 
than  before. 

3.  Solve  each  of  two  equations  for  the  same  unknown  in 
terms  of  the  other  unknowns  and  place  these  two  values  equal 
to  each  other.     The  resulting  equation  will  contain  one  less 
unknown  than  the  two  before. 

For  other  methods  books  on  higher  algebra  or  theory  of 
equations  must  be  consulted. 

1.  Eliminate  c  from  ay  =  x  —  c,  2  ay'y  =  2  (x  —  c). 

2.  Eliminate  c  from  ax  +  by  =  c,  a'x  +  b'y  =  3  c. 

3.  Solve  for  x,  y  from  ax  +  by  =  c,  a'x  +  b'y  =  c'  and  save 
the  result  as  a  formula  for  solving  Ex.  4. 

4.  Apply   the   results   of  Ex.    3,   to   solve  2  x  +  3  y  =  8, 
5x-2y-3. 

5.  Solve  10/s  -  9/y  =  8,  8/x  +  15 /y  =  1. 

Note.  —  Do  not  clear  of  fractions,  but  consider  l/x,  l/y  as 
unknowns. 

6.  Solve  6  x  -  4  y  —X  z  =  17,     9  x  -  7  y  -  16  z  =  29,    10  a; 
-  5  y  -  3  z  =  23. 

7.  If  I  =  Prt  and  A  =  P  (1  +  rt)  and  if  A  =  1250,  r  =  0.06, 
7  =  250,  find  P,  t. 

8.  Solve  a\x  +  biy  +  c\z  =  d\,  azx  +  b2y  +  c&  =  dz,  a$x  +  bsy 
+  c3z  =  d3  and  save  the  result  as  a  formula  to  solve  Ex.  9. 

9.  Apply  the  result  of  Ex.  8,  to  solve  3  x  —  2  y  -}-  z  =  5, 
2s  +  3 y- 4«  =  8,  x-5y  +  3z  =  6. 

10.  Eliminate  x  from  the  equations  xy  =  12,  x  —  y  =  1. 

11.  If  Z  =  a  +  (n  -  1)  d,  S  =  (a  +  1)  n/2,  and  if  a  =  4,  /S  =  60, 
I  =  16,  find  n  and  d. 

12.  If  z  =  19,  y  =  1900;  x  =  25,  y  =  3230;  re  =  44,  y  =  9780 
satisfy  the  equation  y  =  ax2  +  bx  +  c,  determine  a,  6,  c. 

13.  If  a  straight  line  passes  through  the  points  whose  co- 
ordinates are  x  =  2,y  =  3;  x  =  —  4,  i/  =  6  and  if  the  equation 
of  the  line  is  y  =  mx  +  b,  find  m,  b. 


FORMULAS  RELATING  TO  RADICALS  11 

9.  Formulas  relating  to  radicals:    All  ordinary  operations 
with  radicals  are  based  on  the  following: 


2.  a*7'  = 

3.  v^or 
4. 


5.  \~\fa  =  "v/a. 
6. 

7.  < 

8.  6 


9.  vV"  •  v'a"""'  =  a. 

10.  (V^  +  Vb)  (Va  -  Vb)  =  a  -  b. 

These  formulas  are  extensions  of  the  index  laws.    By  means  of 
formula  1  and  the  index  laws  in  article  2  the  remaining  nine 
formulas  are  easily  verified. 
Simplify  the  following: 

1.  VI21.    Thus,  Vl2l  =  VII2  =  11.    See  formula  (1). 

2.  A/625  x^y4^.     Thus,  v/625  xl2yW  =  ^^yz2  =  x  >/5^. 


3.   V480;  VI  (use  formulas  8,  9).     Thus,  V|  =  \/£  x 


4.  ^81  a4/16  be2. 

5.  v/320  -  V^135  +  ^625. 

6.  Vf|  +  V|o  +  \/|. 

Reduce  to  common  index: 

7.  V3,  v/5,  v^. 

Use  formula  (3),  reduce  exponents  to  common  denominator 
and  pass  back  to  radicals. 

*  If  o  is  negative  and  n  an  even  integer,  Va  is  said  to  be  an  imaginary 
quantity.    Such  quantities  are  written  as  follows: 

V^a-  =  V-  1  Vx  =  i  Vx 
where  x  is  positive. 


12  REVIEW  OF  ALGEBRA 

8.  Vl2  +  V75  +  VT47  -  Vi8. 

9.  \/2  a26,  v'G  62c5,  "v/14  c4a7,  to  common  index. 

10.  Vl2  +  \/3  -  V45. 

11.  V3,  "V/5,  VTl,  to  common  index. 

12.  V50  -  ViJ  -  v^24  -  \/ll. 

13.  ViTaF2  •  V8  b*<?  (to  common  index  before  multiplying). 

14.  V^-Vf-V^. 

15.  (Ve  +  VTo  +  vTi)  -5-  V2. 

16.  V5  -  2  \/2  .  V5  -  2  \/2. 


17.    (x  +  y)  V(x  -  y)/x  +  y  -  (x  -  y)  V(x  +  y)/(x  -  y] 


18.  (V3  -  \/2)/(V2  -  V5),  use  formula  (10)  to  remove 
radicals  from  the  denominator. 

19.  (a+  V6)/(o-  Vft). 

20.  (3  +  V6)/(\/2  +  V3). 

Solve  the  following  equations  and  substitute  the  results  in 
the  original  equation  in  each  case. 

21.  x%  =  4,  raise  both  members  to  the  fourth  power. 

22.  (V2x  —  l)^  =  V3,  raise  both  members  to  the  sixth 
power. 

23.  Vx  —  4  +  Vx  —  11  =  7,  square  both  members,  trans- 
pose, square  again. 

24.  l/(VxT~i)  -  l/(Vx~^T)  +  1/Vz2  -  1  =  0,  clear  of 
fractions  and  proceed  as  in  (23). 

25.  V/2  +  \/3  +  Vx  =  2. 

26.  z  -  7  -  Vx2  —  5  =  0,  transpose  x  —  7  before  squaring. 

27.  x*  -  7  -  Vz2  -  4  =  0,  x2  is  to  be  considered  the  un- 
known at  first. 

10.  Each  student  should  be  able  to  solve  quadratic  equa- 
tions quickly  and  accurately  by  some  method.  The  theory  ol 
equations  of  the  second  degree  and  of  higher  degrees  will  be 


A  PAIR  OF  SIMULTANEOUS  EQUATIONS  13 

treated  in  a  later  chapter.  Here  will  be  given  a  formula  which 
will  answer  all  present  needs.  Suppose  the  quadratic  equation 
has  been  reduced  to  the  form 

ax2  +  bx  +  c  =  0. 
Then  the  two  roots  are  given  by  the  formula, 


2a 

A  pair  of  simultaneous  equations  with  two  unknowns,  where 
one  equation  is  of  the  second  degree  and  one  of  the  first  degree, 
may  be  solved  by  the  second  method  of  elimination,  8.  Solve 
the  first  degree  equation  for  one  of  the  unknowns  in  terms  of 
the  other  and  substitute  in  the  second  degree  equation.  The 
resulting  equation  will  be  a  quadratic  with  one  unknown  which 
can  be  solved  by  the  above  formula.  The  other  unknown  can 
then  be  found  by  substituting  in  the  first  degree  equation. 

Solve  the  following  equations: 

1.  z2-2z  =  35. 

2.  z2-3s-l  -  A/3  =  0. 

3.  2x/(x  +  2)  -  (x  +  2)/2z  =  2. 

4.  15z2-86z-64  =  0. 

5.  4  x*  —  17  x2  —  18  =  0,  consider  x2  as  the  unknown  at 
first. 

6.  3**  -4z*  =  7. 

7.  a2  -  1/x*  =  a2  -  I/a2. 

8.  V3  —  x  +  V2  —  x  =  \/5  -  2  x. 
9. 


10>    \2x-Zy  =  5. 

( 1/z3  +  1/0*  =  1001/125.  Divide  first  equation  by  second 
( 1/x  +  l/y  =  11/5.         member  by  member  and  solve 

resulting  equation  with  second. 

x  —  y  =  1. 
13>    \xy  =  (a2  -  62)/4. 


14  REVIEW  OF  ALGEBRA 

11.  Inequalities.  —  Often  the  most  important  lact  about 
two  numbers  is  that  one  is  greater  than  or  less  than  the  other. 
To  express  such  relations  briefly  a  sign  of  inequality  is  used  as 
follows:  a  >  b  means  a  is  greater  algebraically  than  b,  or  that 
b  is  less  than  a. 

If  6  <  a  then  —b  <  —a  «  means  not  less  than),  for  we 
know  that  —  b  >  —a  if  a  >  b. 

The  following  laws  hold  for  inequalities  as  for  equations: 

1.  Both  sides  of  an  inequality  may  be  multiplied  or  divided 
by  the  same  positive  number  without  affecting  the  sense  of 
the  inequality. 

2.  Equal  numbers  may  be  added  to  or  subtracted  from  both 
sides  of  an  inequality.     A  term  may  be  transposed. 

If  both  sides  of  an  inequality  be  multiplied  or  divided  by  the 
same  negative  number  the  ense  of  the  inequality  is  reversed 
and  the  vertex  of  the  sign  must  be  pointed  in  the  opposite 
direction. 

1.  For  what  values  of  x  is  the  expression  7  x  —  23/3  <  2  z/3 
+  5?    Multiply  both  sides  by  3,  21  x  -  23  <  2  x  +  15. 

Transpose,  19  re  <  38. 

Divide  by  19.  x  <  2. 

Therefore  the  inequality  holds  for  all  values  of  x  less  than  2. 

2.  For  what  values  of  x  is 

(x  -  1)  (x  -  2)  (x  -  3)  <  (x  -  5)  0  -  6)  (x  -  7). 

3.  Show  that  for  all  values  of  x,  9  x2  +  25  <  30  x. 

4.  For  what  values  of  x  is  4  x2  —  4  x  —  3  <  0. 

First  determine  for  what  values  of  x  the  expression  is  0. 
Then  by  inspection  determine  the  ones  that  satisfy  the  in- 
equality. 

5.  Show  a/b  +  'b/a  >  2,  for  all  positive  values  of  a,  b  and 
6  7*  a. 

6.  Show  (a  +  6)  (a3  +  63)  >  (a2  -  62)2. 

7.  For  what  values  of  x  is  x  —  7  >  3  x/2  —  8. 

8.  For  what  values  of  x  is  (a;  -  1)  (x  -  3)  (x  -  6)  >  0. 

By  choosing  values  of  x  and  determining  the  signs  of  the 
factors  the  answer  can  be  deduced  without  calculating. 


THE  BINOMIAL  FORMULA  15 

12.  The  binomial  formula  is  of  frequent  use  in  mathematics. 
Let  us  assume: 


(1)    (« 

¥ 
n(n-l)(n  — 2). 


n  (n  -  1)  (n  -  2)  .  .  .  (n  -  r  +  2)  an-f+1fr-1 
1  •  2  •  3    ...  (r  -  1) 


1-2- 3  .  .  .  (r-l)r 
H +  6n. 

Multiplying  both  sides  by  a  +  6  and  carrying  out  the  actual 
multiplication, 

(2)    (a  +  b)n+1  =  an+1  +  (n  +  1)  anb  + ^-= —     .  -|_  .  .  . 

(n  +  l)n  (n  —  1)  .  .  .  (n  —  r  +  3)  an~r+2bf-1 
1-2   ...  (r-1) 

I   (n-rL)n(n—L).  .  .  (n—  y+2)  an   o'  ,  ,  +1 

1.2.  .  .r 

It  is  seen  that  equation  (2)  is  exactly  what  equation  (1)  would 
become  if  in  the  latter  n  +  1  is  put  in  place  of  n.  By  actual 
trial  it  is  known  that  (1)  holds  for  n  —  2.  Then  (2)  gives 
immediately  the  value  of  (a  +  6)3.  Now  in  (1)  put  n  =  3  and 
(2)  gives  the  value  of  (a  +  6)4.  This  process  may  be  continued 
indefinitely.  It  follows  that  (1)  and  (2)  hold  for  any  and  all 
positive  integral  values  of  n.  It  will  be  assumed  here  and 
proved  later  that  (1),  (2)  hold  for  fractional  and  negative  values 
of  n. 

1.  Find  without  multiplying  the  5th  power  of  1  —  x. 

2.  Find  without  multiplying  the  \  power  of  1  —  x  to  4  terms. 

3.  Find  without  multiplying  the  —  \  power  of  1  —  #  to  4  terms. 

4.  Find  without  multiplying  the  —1  power  of  1  —  x  to  4 
terms. 

5.  Find  without  multiplying  the  \  power  of  1  —  x  to  4  terms. 


CHAPTER  II. 
GEOMETRICAL  THEOREMS  AND  FORMULAS* 

13.     1.   A  circle  of  given  radius  can  be  described  about  any 
point  as  a  center. 

2.  In  the  same,  or  in  equal  circles,  equal  central  angles  inter- 
cept equal  arcs. 

3.  The  perimeters  of  inscribed  polygons  are  less  than  the 
circle  and  the  perimeters  of  circumscribed  polygons  are  greater 
than  the  circle. 

4.  If  two  straight  lines  intersect,  the  vertical  angles  formed 
are  equal. 

5.  If  two  parallel  lines  are  cut  by  a  transversal,  the  alternate 
interior  angles  formed  are  equal. 

6.  If  two  parallel  lines  are  cut  by  a  transversal,  the  corre- 
sponding angles  formed  are  equal. 

7.  Angles  having  their  sides  parallel  each  to  each  and  extend- 
ing in  the  same  direction  or  in  opposite  directions  from  the 
vertex  are  equal.     If  one  pair  of  parallel  sides  extend  in  the 
same  direction  and  the  other  pair  in  the  opposite  direction  the 
angles  are  supplementary. 

8.  Angles  having  their  sides  perpendicular  each  to  each,  and 
both  acute  or  both  obtuse,  are  equal. 

9.  The  sum  of  all  the  angles  of  a  triangle  is  equal  to  a  straight 
angle. 

10.  In  any  triangle  there  must  be  at  least  two  acute  angles. 

11.  In  any  isosceles  triangle,  the  angles  opposite  the  equal 
sides  are  equal. 

12.  An  equilateral  triangle  is  equiangular. 

13.  In  an  isosceles  triangle  the  bisector  of  the  vertical  angle 
is  the  perpendicular  bisector  of  the  base. 

*  These  are  given  chiefly  for  reference  use. 
16 


GEOMETRICAL  THEOREMS  AND  FORMULAS  17 

14.  If  two  sides  and  the  included  angle  of  one  triangle  are 
equal  to  the  corresponding  parts  of  another,  the  triangles  are 
congruent. 

15.  If  two  angles  and  a  side  of  one  triangle  are  equal  to  the 
same  parts  of  another,  the  triangles  are  congruent. 

16.  If  three  sides  of  a  triangle  are  equal  to  three  sides  of 
another,  the  triangles  are  congruent. 

17.  Two  lines  perpendicular  to  two  intersecting  lines,  re- 
spectively, must  meet. 

18.  If  two  opposite  sides  of  a  quadrilateral  are  equal  and 
parallel,  the  figure  is  a  parallelogram. 

19.  The  diagonals  of  a  parallelogram  bisect  each  other. 

20.  If  the  diagonals  of  a  parallelogram  are  equal,  the  figure 
is  a  rectangle. 

21.  The  diagonals  of  a  rhombus  bisect  each  other  at  right 
angles. 

22.  The  sum  of  the  interior  angles  of  a  convex  polygon  of  n 
sides  is  n  —  2  straight  angles. 

23.  The  sum  of  the  exterior  angles  of  a  convex  polygon  is  two 
straight  angles. 

24.  Every  point  on  the  perpendicular  bisector  of  a  sect  is 
equidistant  from  the  ends  of  the  sect.     (A  sect  is  a  segment  of  a 
straight  line.) 

25.  Any  straight  line  parallel  to  a  side  of  a  triangle  forms 
with  the  other  two  sides  a  triangle  similar  to  the  first.     (The 
two  sides  produced  if  necessary.) 

26.  A  line  parallel  to  one  side  of  a  triangle  divides  the  other 
two  sides  in  the  same  ratio. 

27.  Homologous  sides  of  similar  triangles  have  the  same  ratio. 

28.  In  any  two  similar  polygons  (1)  the  homologous  angles 
are  equal;   (2)  the  homologous  sides  are  proportional. 

29.  In  any  two  similar  polygons  the  ratio  of  any  two  homol- 
ogous lines  is  equal  to  the  ratio  of  any  other  homologous  lines 
and  equal  to  the  ratio  of  the  perimeters  of  the  polygons. 

30.  The  longer  side  of  a  triangle  is  opposite  the  greater 
angle. 


18  GEOMETRICAL  THEOREMS  AND  FORMULAS 

31.  In  any  triangle  any  side  is  less  than  the  sum  of  the  other 
two  sides  and  greater  than  their  difference. 

32.  The  diameter  of  a  circle  is  twice  its  radius. 

33.  A  circle  is  determined  by  (a)  the  center  and  radius;   (6) 
center  and  diameter;   (c)  three  points  not  in  a  straight  line. 

34.  A  perpendicular  bisector  of  a  chord  passes  through  the 
center  of  the  circle. 

35.  A  straight  line  perpendicular  to  a  radius  at  its  extremity 
is  tangent  to  the  circle. 

36.  Equal  chords  are  equally  distant  from  the  center  of  the 
circle. 

37.  Two  angles  at  the  center  have  the  same  ratio  as  their  in- 
tercepted arcs. 

38.  The  area  of  a  rectangle  or  a  parallelogram  equals  the 
product  of  the  base  by  the  altitude. 

39.  The  area  of  a  triangle  equals  the  product  of  the  base  by 
half  the  altitude. 

40.  The  area  of  a  circle  equals  IT  times  the  square  of  its 
radius,  (ur2).     TT  =  3.1416  approximately. 

41.  The  area  of  a  sphere  equals  4?r  times  the  square  of  its 
radius,  (4  irrz). 

42.  The  sum  of  the  squares  of  the  legs  of  a  right  triangle 
equals  the  square  of  the  hypotenuse. 

43.  The  altitude  of  a  right  triangle  from  the  right  angle  is  a 
mean  proportional  between  the  segments  into  which  it  divides 
the  hypotenuse. 

44.  The  volume  of  a  prism  or  cylinder  equals  the  area  of  its 
base  times  its  altitude. 

45.  The  areas  of  similar  polygons  have  the  same  ratio  as  the 
squares  on  any  two  homologous  lines. 

46.  Area  of  triangle  whose  sides  are  a,  b,  c,  is  given  by 


A  =  Vs  (s  —  a)  (s  —  6)  (s  —  c), 
when  2s  =  a  +  6  +  c. 

47.   Area  of  sector  of  circle  A  =  — ~ — • 


GEOMETRICAL  THEOREMS  AND  FORMULAS  19 

48.  Area  of  segment  of  sphere  A  =  2irrh  where  h  is  altitude 
of  segment. 

49.  Area  of  curved  surface  of  cylinder,  A  =  2  irrh. 

50.  Area  of  curved  surface  of  cone,  A  =  2  irr  ~ ,  where  s  is 

£t 

slant  height. 

51.  Area  of  a  segment  (of  one  base)  of  a  circle  is  the  area  of 
the  sector  subtended  by  the  arc  of  the  segment  minus  the  area 
of  the  triangle  whose  base  is  the  chord  of  the  segment  and 
vertex  the  center  of  the  circle. 


CHAPTER  III 
METHODS   OF  CALCULATION 

14.  The  need  of  numbers  and  number  calculations  arose 
quite  naturally  in  race  development.     Recognition  of  number 
and  geometric  form  appears  not  to  be  peculiar  to  the  human 
race.     There  is  good  reason  for  believing  that  many  of  the  lower 
races  of  animals  have  the  ability  to  count  and  to  recognize  space 
forms  and  magnitudes.     A  few  of  the  conditions  in  human 
affairs  that  called  forth  methods  of  counting,  measuring  and 
reckoning  are  enumerated  below. 

1.  Numbering  and  comparing  groups  of  objects. 

2.  Barter  and  trade  of  primitive  commerce. 

3.  Census  and  tribal  statistics. 

4.  Calendar  and  time  measurements. 

5.  Land  measurements. 

6.  Determination  of  weights  and  measures. 

7.  Taxation. 

8.  Modern  commerce. 

9.  Engineering  and  exact  science. 

10.  Personal  and  corporation  accounting. 

11.  Insurance,  savings  accounts  and  investments. 

12.  Statistical  and  social  science. 

15.  Important  discoveries  in  the  art  of  calculating.  —  (a) 
-Hindu   notation:     We   owe   our  present   method   of  writing 
ordinary  numbers  to  the  Hindus.     The  important  principles  of 
this  notation  and  the  ones  that  make  it  superior  to  all  others, 
to  date,  are  its  position  value  and  zero.     This  notation  probably 
antedates  the  Hindus  but  it  was  through  them  that  it  was 
transmitted  to  the  western  nations. 

20 


MECHANICAL  CONTRIVANCES  21 

(6)  Decimal  fractions  were  given  to  the  world  about  the  end 
of  the  sixteenth  century  by  Simon  Stevins  of  Bruges  in  Belgium. 
Fractions  had  been  a  hard  problem  for  the  race.  For  a  long 
time  all  fractions  were  expressed  as  the  sums  of  unit  fractions. 
Thus  7/12  was  regarded  as  1/2  +  1/12  and  similarly  for  other 
fractions.  Other  methods  were  developed  gradually.  The 
discovery  that  fractions  could  be  incorporated  into  the  number 
system  in  decimal  form  as  an  extension  of  the  Hindu  notation 
was  a  real  contribution  not  only  to  mathematics  but  to  civili- 
zation. 

(c)  Early  in  the  seventeenth  century  the  discovery  of  loga- 
rithms gave  the  computer  and  mathematician  another  power- 
ful instrument.  It  was  the  crowning  achievement  in  numbers. 
Coming  at  a  time  when  Kepler  was  working  on  his  planetary 
theory  and  immediately  following  the  computation  of  trigono- 
metric tables  by  the  Germans,  the  discovery  of  logarithms  was 
of  great  value.  The  names  of  Napier,  the  inventor,  and  Briggs, 
the  collaborator  and  editor,  of  the  tables  have  been  made 
immortal  by  these  contributions.  Laplace  remarked  that  the 
discovery  of  logarithms  would  double  the  life  of  the  astronomer 
by  shortening  the  labor  of  his  calculations. 

16.  Many  mechanical  contrivances  have  been  and  are  still 
in  use  for  shortening  the  labor  of  computing.  In  passing  we 
may  mention :  (a)  The  abacus,  an  ancient  instrument,  still  in 
use  in  some  countries  and  used  in  our  schools  under  the  name, 
numeral  frame;  (6)  the  slide  rule,  which  is  based  on  the  idea 
of  logarithms.  It  is  convenient  and  adequate  for  many  purposes 
in  applied  science  and  in  commerce;  (c)  the  calculating  machine, 
such  as  the  Burroughs,  does  all  the  ordinary  arithmetical 
calculations  with  a  speed  and  accuracy  that  justifies  its  use  in 
banks  and  mercantile  houses  where  much  computing  is  done; 
(d)  geometrical  methods  of  performing  the  arithmetical  opera- 
tions were  in  possession  of  the  Greeks.  In  certain  branches  of 
engineering,  geometric  methods  are  now  used  with  great 
efficiency.  They  depend  on  the  principle  of  the  proportionality 
of  the  sides  of  similar  triangles,  (e)  General  reckoning  tables, 


22  METHODS  OF  CALCULATION 

containing  products,  quotients,  powers,  roots  and  reciprocals  of 
numbers  may  be  obtained  by  those  who  desire  to  use  them. 

17.  Graphic  or  geometric  representation  of  numbers.  - 
A  number  is  graphically  represented  by  a  straight  line  of  length 
proportional  to  the  magnitude  of  the  number.  The  factor  of 
proportionality  determines  the  scale  of  the  representation. 
Thus,  if  a  line  1"  *  long  is  to  represent  the  number  50,  the  scale 
is  50  to  1  in  inches.  If  the  number  275  is  to  be  represented 
by  a  line  not  over  4"  long,  we  divide  275  by  4  and  obtain  69, 
nearly.  It  would,  therefore,  be  safe  to  use  a  scale  of  70  to  1  or 
any  larger  number  than  70  to  1. 

It  must  be  remembered  that  the  smaller  the  scale  (larger 
ratio  of  proportionality),  the  more  difficult  it  is  to  estimate 
small  numbers  or  odd  units.  On  the  other  hand  the  larger  the 
scale  (smaller  ratio  of  proportionality),  the  larger  the  drawing 
and  the  larger  the  paper  or  other  surface  required.  The  scale 
is  to  be  determined  by  convenience  and  by  the  degree  of 
accuracy  demanded.  Very  large  drawings  are  sometimes  nec- 
essary. Sometimes  very  small  drawings  will  do. 

1.  (a)  Lay  off  350  to  the  scale  of  50  to  1";   (6)  lay  off  350  to 
the  scale  of  100  to  1". 

2.  Measure  the  lines  below  to  the  scale  of  50  to  1"  and  deter- 
mine the  numbers  they  represent. 

A 


FIG.  1. 

18.  Arithmetical  operations.  —  (1)  To  add  a  number  6  to 
a  number  a:  Let  01  be  the  unit  length.  Suppose  a  =  OA, 
b  =  AB.  Obviously  OB  =  OA  +  AB  represents  the  sum  of 
a  and  6,  to  the  unit  01. 


B 
FIG.  2. 


*  The  notation,  1",  means  1  inch;  1'  means  1  foot. 


ARITHMETICAL  OPERATIONS  23 

To  subtract  b  from  a  lay  off  6  from  A  toward  B'.  Then 
a  -  b  =  OA  -  AB  =  OB'. 

(2)  To  multiply  a  number  a  by  a  number  6.  Let  01  be  the 
unit.  Lay  off  a  =  OA.  From  A  draw  a  line  AC  making  any 
convenient  angle  with  OA.  Lay  ofi  IB  parallel  to  AC  and 
equal  to  b.  Draw  OB  and  produce  it  until  it  intersects  AC  at 
C,  say.  Then  AC  represents  the  product  ab.  That  is,  c  = 
ab.  For,  by  the  similar  triangles  OAC,  01B, 

01  :  IB  :  :  OA  :  AC 

or 

01  -  AC  =  IB  -  OA 
and 

AC  =  IB  •  OA, 

c  =  ba, 
since  01  =  1,  the  unit  of  measure. 


FIG.  3. 

(3)  To  divide  a  number  c  by  a  number  a.  In  this  problem 
use  the  diagram  of  the  preceding  problem  and  solve  the  last 
equation  for  b.  It  is  seen  that  if  a  is  the  divisor  and  c  the  divi- 
dend, the  quotient  is  6. 

The  reciprocal  of  a  number  can  be  found  as  a  special  case 
where  the  dividend  is  1.  In  using  this  method  for  finding 


24 


METHODS  OF  CALCULATION 


reciprocals,  it  is  advisable  to  employ  a  larger  scale  for  the 
dividend  than  for  the  divisor. 

The  square  of  a  number  can  be  found  as  a  special  case  of 
problem  (2)  where  a  =  b. 

(4)  To  find  the  square  root  of  a  number  a:  Evidently  a  = 
a  •  1  and  Va  =  VoTI.  Make  BC  =  a  and  AB  =  1.  Draw  on 
AC  as  a  diameter,  a  circumference,  AKC.  At  B  erect  a  per- 
pendicular  meeting  the  circumference  at  D.  Then  BDZ  = 
AB  -  BC  by  geometry.  Hence  Va  =  5D. 


— >c 


1.  Find  the  product  of  35  X  73;   18  X  93;  36  X  13  by  the 
geometric  method. 

2.  Find  the  quotient  of  125/16;   39/4;   50/13  by  the  geo- 
metric method. 

3.  Find  the  reciprocals  of  all  integers  from  2  to  10,  inclusive, 
by  the  geometric  method. 

4.  By  the  geometric  method  find  Vl2;  V27;  Vl8;  V56. 

5.  By  the  geometric  method  find  (35  X  12)/16;   18x27/13. 

19.  Idea  of  logarithms.  —  The  student  should  now  recall 
the  index  laws,  see  2,  Chap.  I. 

All  positive  numbers  can  be  regarded  as  powers  of  some 
positive  number,  not  unity.  It  is,  of  course,  true  tkat  most  such 
powers  cannot  be  exactly  expressed,  but  close  approximations 
can  be  determined  and  exactly  expressed.  This  is  because  these 
indices  are  not  rational  numbers.  A  rational  number  can  be 
expressed  as  the  quotient  of  two  integers. 


RULES  25 

What  power  of    2  is    8? 
What  power  of    2  is  10? 
What  power  of  10  is  10? 
What  power  of  10  is  100? 
What  power  of  10  is  1/10? 

20.  Definition  of  the  logarithm  of  a  number.  —  If  y  =  ax, 
x  is  called  the  logarithm  of  y  to  the  base  a.     Briefly  we  may 
write  x  =  loga  y.    y  =  ax  implies  x  =  loga  y  and  conversely. 

From  this  definition  it  follows  that 

logiolO  =  1,  since  10  =  101, 
logio  100  =  2,  since  100  =  102, 
logic  1/100  =  -2,  since  1/100  =  10~2. 

21.  Rules    for    calculating    with    logarithms    are    derived 
directly  from  the  index  laws.     Thus, 

102  •  101  =  102+1  =  103  =  1000. 
That  is; 

(1)  logic  (102  •  10)  =  log™  1000  =  3  =  2  +  1  =  logio  100  +  logio  10. 
Again : 

(2)  logio  (1000 ^-10)=  logio  100  =  2  =  3-l=logio  1000-logio  10. 

Generalized,  (1)  and  (2)  give  the  rules  for  multiplication  and 
division  by  means  of  logarithms.  Let  x  and  y  be  any  two 
positive  numbers  and  p  their  product.  Then, 

(!')  loga  p  =  loga  x  +  logo  y. 

• 
If  q  is  the  quotient  x/y,  then 

(2')  loga  q  =  loga  X  -  loga  y. 

Consider  (3)  y  =  xn,  where  n  is  any  number.     Then,  by  20 
loga  y  =  logo  yory  =  a10**,    x  =  a10*-1,    xn  =  (a108*1)"  =  a*10*-1. 
Therefore, 
(3')  loga  y  =  loga  xn  =  n  loga  x. 

This  result  shows  how  to  use  logarithms  to  raise  a  number  to 
any  power. 


26  METHODS  OF  CALCULATION 

Now  consider: 

i 
(4)  y  =  Vz  =  xm. 

We  can  write: 

L        I 

(40  10ga  y  =  bga  £OT  =  —  loga  X. 

Ub 

This  result  shows  how  to  use  logarithms  to  find  the  roots  of 
numbers. 

The  foregoing  equations  (I'),  (2'),  (30,  (40  may  now  be 
translated  into  rules. 

1.  The  logarithm  of  the  product  of  two  or  more  numbers  is 
the  sum  of  the  logarithms  of  the  numbers. 

2.  The  logarithm  of  the  quotient  of  one  number  divided  by 
another  is  the  logarithm  of  the  dividend  minus  the  logarithm  of 
the  divisor. 

3.  The  logarithm  of  the  nth  power  of  a  number  is  n  tunes  the 
logarithm  of  the  number. 

4.  The  logarithm  of  the  nth  root  of  a  number  is  the  logarithm 
of  the  number  divided  by  n. 

5.  The  logarithm  of  the  reciprocal  of  a  number  is  the  negative 
of  the  logarithm  of  the  number.     (See  co-logarithm  below.) 

It  may  be  more  convenient  to  use  addition  when  dividing 
with  logarithms.  To  do  this  the  logarithm  may  be  subtracted 
from  0  or  10  —  10,  or  20  —  20,  etc.  Thus  instead  of  subtract- 
ing log  n  =  3.37581,  it  may  be  more  convenient  to  use  addition. 
This  is  easily  done  by  first  subtracting  3.37581  from  10  —  10. 
Thus  we  may  add  6.62419  —  10.  The  advantage  of  this  method 
lies  in  the  fact  that  it  is  easy  to  make  the  above  subtraction 
directly  as  the  logarithm  is  taken  from  the  table  by  subtracting 
each  figure,  beginning  at  the  left,  from  9  and  the  last  figure  on 
the  right  from  10.  The  result  of  this  subtraction  is  called  the 
co-logarithm  of  the  number  or  arithmetic  complement  of  the 
logarithm. 

22.  From  19  and  20  it  is  evident  that  when  10  is  the  base 
of  the  logarithms,  the  following  statements  hold : 


RULES  27 

1.  Any  number  between  1  and  10  has  a  proper  fraction  for 
its  logarithm. 

2.  Any  number  between  10  and  100  has  1  plus  a  proper 
fraction  for  its  logarithm. 

3.  Any  number  between  100  and  1000  has  2  plus  a  proper 
fraction  for  its  logarithm. 

4.  Any  number  between  0  and  1  has  a  negative  logarithm. 
It  is  seen  that  logarithms  of  numbers  consist  of  two  parts,  an 

integer  or  zero  and  a  fraction.  The  integer  part  is  called  the 
characteristic.  The  fractional  part  is  called  the  mantissa  of 
the  logarithm. 

For  convenience  in  calculating,  a  logarithm  should  be  written 
so  the  part  on  left  of  decimal  point  is  positive.  The  mantissa  is 
always  positive.  Thus,  2.34567  (where  2  shows  that  the  char- 
acteristic, only,  is  negative)  may  be  expressed  as  8.34567  —  10 
where  the  part  of  the  characteristic  on  left  of  the  decimal  point 
is  positive. 

Consider: 

log  273  =  2  plus  a  proper  fraction. 

log  27.3  =  1  plus  the  same  proper  fraction. 

log  2.73  =  0  plus  the  same  proper  fraction. 

log  0.273  =  I  plus  the  same  proper  fraction. 

log  0.0273  =  2  plus  the  same  proper  fraction. 

These  results  are  evident  from  the  fact  that,  when  the  decimal 
point  is  moved  one  place  to  the  left,  the  number  is  divided  by 
10  and  exactly  1  is  subtracted  from  its  logarithm,  the  mantissa 
remaining  the  same.  It  is  easy  to  see  that  the  characteristic 
depends  on  the  position  of  the  decimal  point  in  the  number, 
while  the  mantissa  depends  on  the  sequence  of  digits  in  the 
number.  It  will  be  noticed  that  with  a  number  =  1  the  char- 
acteristic is  an  integer  one  less  than  the  number  of  significant 
figures  of  the  number  to  the  left  of  the  decimal  point.  This  rule 
is  general  if  the  number  of  O's  immediately  to  the  right  of  the 
decimal  point  be  considered  as  a  negative  number  of  figures  on 
the  left.  Note  that  the  above  example  verifies  this  statement. 


28  METHODS  OF  CALCULATION 

What  is  the  characteristic  of  the  logarithm  of  each  of  the 
following  numbers:  3.047,  37.56,0.000842,  1.0045,  67,543,0.43? 

23.  In  order  to  use  logarithms  in  calculating  it  is  necessary 
to  have  a  table  of  logarithms  of  all  numbers  used  in  the  cal- 
culations.     The  mantissas  only  are  given  in  the  table.     The 
characteristic  must  be  supplied  in  every  case  according  to  the 
above  rule. 

To  construct  a  table  of  logarithms  involves  great  labor.  It  is 
beyond  the  scope  of  this  course  to  explain  the  method  of  cal- 
culating such  tables.  For  this  information  the  student  is 
referred  to  works  on  higher  algebra  or  calculus.  A  series  from 
which  logarithms  may  be  calculated  will  be  given  in  Chap.  XVI. 
A  four  place  table  of  logarithms  of  numbers  with  explanations 
is  to  be  found  in  the  back  of  the  text. 

Solve  the  following  by  the  use  of  logarithms. 

1.   79  •  470  •  0.982. 

By  the  rule  for  multiplication  of  21  write 

log  (79  •  470  •  0.982)  =  log.79  +  log  470  +  log  0.982 
=  1.8976 
+2.6721 
+9.9921  -10  =  1.9921 

=  14.5618  -  10  =  4.5618  =  log 3646. 

2.  9503  •  0.7095  =  ?  5.   (2.58S)5  =  ? 

3.  8075  -4-  364.9  =  ?  6.   (0.57)-4  =  ? 

4.  (-0.643) .  0.7564  =  ?         7.   ^-0.3089  =  ? 

8.  (0.000684)*  =  ? 

9.  V943  •  (-7298)7(0.00006.  (-9~9))  =  ? 
10.    (\/0.0476  '  ^222)7  ^5059  •  0.0884  =  ? 

24.  An  exponential  equation  is  one  in  which  the  unknown 
quantity  occurs  as  the  exponent  of  some  known  number  in  the 
equation.     Such  equations  can  often  be  easily  solved  by  the 
use  of  logarithms. 


AN  EXPONENTIAL  EQUATION 


29 


Find  the  value  of  x  in  the  equation  132*  =  14r+1.     Taking 
logarithms  of  both  sides, 

2zlogl3  =  (x+l)log!4 
and 

,„  log  14 

2 log  13 -log  14* 

Substituting  the  logarithms  of  13  and  14  gives  x  =  1 .06. 
Use  logarithms  in  solving  the  following  problems. 

1.  Given  A  =  P  (1.0  r)',  I  =  A  —  P,  find  the  compound 
interest  of  $250  at  5  per  cent  for  25  years.  t  (P  =  principal, 
r  =  rate,  t  =  time,  A  =  amount.) 

Note.  —  .Or  may  be  written  r/100  if  preferred. 

2.  In  how  many  years  will  $5000  amount  to  $6000,  com- 
pounded annually  at  6  per  cent?     Given  A  =  P  (1.0  r)*. 

3.  Find  the  value  of  x  in  5*+5  =  8I+1. 

4.  Find  the  value  of  k  in  Ps  =  P<fTk*,  if  P0  =  14.72,  z  = 
1122,  P,  =  14.11,  e  =  2.718. 

5.  How  many  gallons  in  a  cylindrical  tank  3'  in  diameter 
and  6' 8"  high? 

Note.  —  6'  =  6  ft.;   8"  =  8  in.     This  notation  will  be  used 
from  here  on. 

6.  If  c  =  (Ld/fl6)   (262  -  n2),  find  c  when  L  =  1650,  b  = 
500,  n  =  25,  R  =  1000,  d  =  0.75. 

7.  Compute  the  simple  interest  of  $135.70  at  3|  per  cent  for 
12$  yrs.     (7  =  PH.) 

8.  In  the  triangle  ABC,  a  =  175,  6  =  225,  c  =  190.     Find 
the  area.    (A  =  Vs(s  —  a)  (s  —  6)  (s  —  c)  and  s  =  (a  +  b  +  c)/2. 

9.  Find  the  cost  of  covering  a  floor  like  the  figure  at  $1.75 
per  sq.  yd. 


<  t              r,o'             ^ 

^s 

*bo 

I 

..11'. 
> 

FIG.  5. 


30  METHODS  OF  CALCULATION 

10.  Find  the  number  of  gallons  of  paint  required  to  paint  a 
cylindrical  tank  10'  in  diameter  and  40'  long  (curved  surface 
and  both  ends  to  be  painted),  if  1  gal.  paint  is  sufficient  to  paint 
100  sq.  ft.  of  surface. 

11.  A  grindstone  will  stand  a  rim  speed  of  2500'  per  min. 
How  many  r.p.m.  (revolutions  per  min.)  will  a  stone  42"  in 
diameter  stand? 

12.  If  the  cutting  speed  of  a  lathe  is  40'  per  min.,  how  many 
r.p.m.  must  a  lathe  have  to  give  the  desired  speed  for  cutting 
a  piece  2"  in  diameter? 

13.  Find  the  number  of  barrels  capacity  of  a  rectangular 
cistern  6'  X  8'  X  10'.     (7.48  gal.  =  1  cu.  ft.) 

14.  A  pie  is  10"  in  diameter.     It  is  cut  in  6  equal  sectors 
about  the  center.    What  is  the  area  of  the  upper  surface  of 
each  piece? 

15.  Find  the  value  of  a  bin  of  wheat  8'  8"  X  10'  6"  X  4'  3" 
at  $1.95  per  bu. 

25.  The  logarithmic  scale.  —  If  distances  proportional  to 
the  logarithms  of  numbers  are  laid  off  from  one  end  of  a  straight 
line  segment  AB,  the  segment  so  divided  is  a  logarithmic  scale. 


1,  2  3  456789  10 

A  B' 

FIG.  6. 

If  a  duplicate  scale  A  'B'  is  laid  along  side  AB  so  as  to  slide 
parallel  to  itself,  we  can  use  these  scales  to  perform  the  opera- 
tions of  multiplication  and  division  of  numbers.  For  example, 
to  multiply  25  X  30,  set  A'  at  25  (between  2  and  3)  on  AB. 
Run  over  on  A  'B'  to  30  (at  3  on  A  'B')  and  on  A  B  opposite  this 
read  75  (between  7  and  8)  on  AB.  This  gives  the  figures  of  the 
product.  Knowing  the  orders  of  the  numbers  we  know  the 
product  is  750. 

Since  the  scales  are  proportional  to  the  logarithms  of  numbers, 
what  has  been  done  is  to  add  the  logarithm  of  30  to  the  loga- 
rithm of  25  to  obtain  the  logarithm  of  750.  An  instrument 
operating  in  this  way  is  called  a  slide  rule.  Directions  for  use 


31 

generally  accompany  the  rule.  The  student  should  now  obtain 
a  slide  rule  and  learn  to  use  it  as  an  economy  and  as  a  check  in 
his  calculations.  Commercial  and  engineering  concerns  employ 
the  slide  rule  for  certain  types  of  work  with  great  advantage. 

26.  For  convenience  of  reference,  rules  for  the  ordinary 
Manheim  slide  rule  are  appended  here. 

Multiplication.  —  Set  the  index  of  C  scale  over  multiplicand 
on  D  scale.  Under  the  multiplier  on  C  scale  read  product  on 
D  scale. 

Division.  —  Above  dividend  on  D  scale  set  divisor  on  C  scale. 
Under  index  of  C  scale  read  quotient  on  D  scale. 

Proportion.  —  Set  first  term  on  C  scale  over  second  term  on 
D  scale.  Under  third  term  on  C  scale  read  fourth  term  on 
D  scale. 

Squares.  —  Set  the  runner  on  the  number  on  D  scale.  Read 
square  under  runner  on  A  scale. 

Square  root.  —  If  number  has  an  odd  number  of  integral 
digits  use  left  half  of  A  scale;  if  an  even  number  of  integral 
digits  use  right  half  of  A  scale.  Set  runner  over  number  on  A 
scale.  Read  square  root  under  runner  on  D  scale. 

Cubes.  —  Set  index  of  B  scale  under  number  on  A  scale. 
Read  cube  on  A  scale  over  number  on  C  scale. 

Cube  root.  —  Under  number  on  A  scale  move  slide  until 
number  on  A  scale  above  index  of  B  scale  is  same  as  appears 
on  C  scale  under  given  number  on  A  scale. 

Exercise.  —  Solve  all  problems  of  18  by  use  of  the  slide  rule. 

26a.  Methods  of  Interpolation.  —  Immediately  preceding 
the  tables  in  the  back  of  the  text  is  an  explanation  of  the  use 
of  those  tables.  A  number  of  problems  are  worked  which 
illustrate  simple  interpolation. 

Below  are  given  problems  which  illustrate  "double  in- 
terpolation." Other  types  of  interpolation  are  given  in 
Chapters  IV  and  VIII. 

Following  is  a  section  of  a  table  given  in  the  (J.  S.  Weather 
Bureau  Bulletin  No.  235.  A  similar  table  is  used  in  connec- 
tion with  work  in  artillery  and  in  the  navy. 


32 


TABLE.  —  RELATIVE  HUMIDITY,  PER  CENT  —  FAHRENHEIT 

TEMPERATURES 
Pressure  =  29.0  Inches 


Depression  of  wet-bulb  thermometer  (t-tr). 

Air 

temp. 

/. 

0.5 

1.0 

1.5 

2.0 

2.5 

3.0 

3.5 

4.0 

4.5 

5.0 

5.5 

6.0 

6.5 

7.0 

7.5 

8.0 

8.5 

9.0 

9.5 

10.0 

40° 

96 

92 

88 

84 

80 

76 

72 

68 

64 

61 

57 

53 

49 

46 

42 

38 

35 

31 

27 

23 

41 

96 

92 

88 

84 

SO 

77 

73 

69 

65 

62 

58 

54 

50 

47 

43 

40 

36 

33 

29 

26 

42 

96 

92 

88 

85 

81 

77 

73 

70 

66 

62 

59 

55 

51 

48 

45 

41 

38 

34 

31 

28 

43 

96 

92 

88 

85 

81 

78 

74 

70 

67 

63 

60 

56 

52 

49 

46 

43 

39 

36 

32 

29 

44 

96 

93 

89 

85 

82 

78 

74 

71 

68 

64 

61 

57 

54 

51 

47 

44 

40 

37 

34 

31 

45 

96 

93 

89 

86 

82 

79 

75 

71 

68 

65 

61 

58 

55 

52 

48 

45 

42 

39 

36 

33 

Example:  Air  temperature  t  =  42.7°. 

Reading  of  wet-bulb  thermometer  t'  =  34.5°. 

Depression  of  wet-bulb  thermometer  (t  —  t'}  =  8.2°. 
Determine  the  relative  humidity  from  above  table. 

Tabulated  values  are : 


8.0 

8.5 

42 

41 

38 

43 

43 

39 

To  a  difference  of  1°  in  air  temperature  (43-42)  corresponds 
a  difference  of  2  per  cent  in  the  relative  humidity  (43-41) 
when  depression  is  8.0°.  Therefore,  a  change  of  0.7°  in  air 
temperature  (42.7  —  42)  causes  a  change  of  0.7  X  2  per  cent 
=  1.4  per  cent  in  relative  humidity.  Similarly,  for  a  depres- 
sion 8.5°,  a  difference  of  1°  in  air  temperature  causes  a  differ- 
ence of  1  per  cent  in  relative  humidity,  and,  therefore,  a  change 
of  0.7°  causes  a  change  of  0.7  per  cent  in  the  relative  humidity. 
Our  table  may  now  be  enlarged  to 


8.0 

8.5 

42 

41 

38 

42.7 

42.4 

38.7 

43 

43 

39 

METHODS  OF  INTERPOLATION 


33 


When  the  depression  is  8.2°,  the  relative  humidity  by  the 
usual  method  of  interpolation  is: 

Difference  of  .5°  depression  causes  a  difference  of  3.7  per  cent 

in  relative  humidity  (42.4  —  38.7).     Therefore,  difference  of 

0.2°  depression  causes  a  difference  of  1.5  per  cent  in  relative 

humidity.     The  required  humidity  is  therefore  (42.4  per  cent 

-  1.5  per  cent)  =  40.9  per  cent. 

1.  Work  the  above  illustrative  problem  by  changing  the 
order  of  the  method,  that  is,  interpolate  for  temperature  42° 
and  depression  8.2°,  which  lies  between  8.0°  and  8.5°.     Then 
do  likewise  for  temperature  43°  and  so  forth. 

2.  If  air  temperature,  t  =  44.3°, 

Reading  of  wet-bulb  thermometer  t'  =  34.5°, 
Depression  of  wet-bulb  thermometer  (t  —  t')  =  9.8°, 
Determine  the  per  cent  humidity. 

3.  Temperature  is  41.8°, 

Depression  of  wet-bulb  thermometer  (t  —  t'}  —  2.4°, 

Determine  the  per  cent  humidity. 

The  motion  of  a  projectile  is  retarded  by  the  resistance  of 
the  air.  The  amount  of  retardation  due  to  this  cause  depends 
upon  the  temperature  and  barometric  pressure.  Following  is 
a  portion  of  a  ballistic  table  from  which  is  determined  an 
"atmospheric  factor"  that  enters  into  the  computation  of 
retardation1 

BALLISTIC  TABLE 


Temperature  of  air  —  Fahrenheit  degrees. 

30 

31 

32 

33 

34 

35 

36 

37 

38 

28" 

0.994 

0.996 

0.998 

1.000 

1.003 

1.005 

1.007 

1.009 

1.011 

29 

0.960 

0.962 

0.964 

0.966 

0.968 

0.970 

0.972 

0.974 

0.976 

30 

0.928 

0.930 

0.932 

0.934 

0.936 

0.938 

0.940 

0.943 

0.945 

31 

0.898 

0.899 

0.902 

0.903 

0.906 

0.907 

0.909 

0.911 

0.913 

1.  Given  the  temperature  t  =  32°  and  the  barometric  pres- 
sure p  =  29",  then  the  corresponding  atmospheric  factor  is 
0.964.  Verify  by  use  of  the  table. 


34  METHODS  OF  CALCULATION 

2.  Given  the   temperature   t  —  31.8°   and   the   barometric 
pressure    p  =  28.4".     Find    the    corresponding    atmospheric 
factor. 

Note.  —  Carry  out  the  interpolations  in  the  same  manner  as 
was  done  in  the  illustrative  problem  above  on  humidity  of  the 
air. 

3.  If  t  =  31°  and  p  =  30.8",  find  the  corresponding  atmos- 
pheric factor. 

4.  If  t  =  36.6°  and  p  =  30.2",  find  the  atmospheric  factor. 

SUPPLEMENTARY  EXERCISES 

1.  The  diagonal  of  a  square  is  73.84'.     Find  the  length  of  a  side. 

2.  The  circumference  of  a  circle  is  63.24'.     Find  the  radius.     (Use  3.142 
for  TT.) 

3.  If  the  area  of  a  circle  is  given  by  the  formula  A  =  0.7854  d2,  find  the 
radius  of  a  circle  whose  area  is  842.3  sq.  ft. 

4.  If  1"  =  2.540  cm.,  how  many  centimeters  in  8Ty? 

5.  If  the  barometer  reading  is  29.92",  what  will  a  barometer  read  that 
is  graduated  in  millimeters?     If  the  barometer  reading  is  75.46  cm.,  what 
will  a  barometer  read  that  is  graduated  in  inches? 

6.  To  find  the  reciprocal  of  a  number  by  the  use  of  logarithms:   Sub- 
tract the  mantissa  of  the  log  of  the  number  from  1,  add  1  to  its  character- 
istic and  change  the  sign.     The  number  whose  log  :s  this  result  is  the 
reciprocal  of  the  given  number.     Use  this  rule  in  finding  the  reciprocal  of 
543.     Find  log  1/543  by  use  of  a  co-logarithm.     How  do  the  two  methods 
for  finding  the  reciprocal  of  a  number  differ  ? 

7.  Find  the  reciprocal  of  0.00635  by  the  two  methods  of  the  above 
problem. 

8.  If  1  cu.  ft.  of  water  weighs  62.45  Ibs.,  what  is  the  pressure  per  sq.  in. 
of  a  column  of  water  1  ft.  high? 

9.  The  volume  of  a  sphere  v  =  4/3  irr3.    Show  that  r  =  (  j—  j  and  find 

r  in  cm.  when  v  =  37.45  cu.  in. 

10.  The  time  required  for  a  simple  pendulum  to  make  a  single  oscilla- 
tion when  the  angle  through  which  it  swings  is  small,  is  expressed  by 
t  =  IT  VT/0,  where  I  represents  the  length  of  the  pendulum  and  g  represents 
the  acceleration  of  gravity.    What  is  the  length  of  a  pendulum  that  vi- 
brates seconds  when  g  =  979.4  cm. /sec2? 

11.  The  equation  of  motion  of  a  body  falling  from  rest  under  the  action 
of  gravity  is  s  =  f  gt2  where  s  =  distance  and  t  =  tune. 


METHODS  OF  INTERPOLATION  35 

(a)  If  g  =  32.16  ft./sec2  and  s  =  164.8  ft.,  find  t. 

(b)  If  g  =  980.3  cm./sec2  and  s  =  50.23  m.,  find  t. 

Note  that  g  is  expressed  in  centimeters  and  s  in  meters.    Must  this  be  taken  into  con- 
sideration when  computing  t  ? 

12.  If  a  250  lb.  charge  of  a  certain  lot  of  nitro-cellulose  powder  at  70°  F. 
gives  a  muzzle  velocity  of  2275  ft./sec.,  what  should  be  the  weight  of  a 
charge  to  give  a  muzzle  velocity  of  2750  ft./sec.?     Muzzle  velocity  and 

V       f  w\v 
weight  of  charge  are  connected  by  the  formula  =~-  =  f  —  1  .   y  =  1.2  for 

nitro-cellulose  powder. 

13.  Using  same  formula  as  in  problem  (12)  find  the  muzzle  velocity  for 
a  charge  of  275  Ibs.  where  the  velocity  is  2150  ft./sec.  at  30°  F.,  for  a  charge 
of  240  Ibs.  of  nitro-glycerine  powder,     y  =  0.8  for  nitro-glycerine  powder. 

14.  The  following  formulas  are  used  for  range  correction  in  gunnery: 

r  -f     c    f  -1  i  2W*T* 

wl   — Jw  '  t-'j     Jw  —  *  "I  Y 

Find  d  when  T  =  36.60,  Wx  =  25.00,  X  =  54,000,  and  C  =  10.49. 

16.    Given  (J)n  =  J;  find  n. 

Note.  —  Take  the  reciprocal  of  each  member. 

16.  Given  (l/2.641)n  =  1/(159.4);  find  n. 

17.  Given  (l/(0.0649)n  =  1/2.54;  find  n. 

n—l 

rVA*""1         /Po\  n 
^  I       =  (  TT  J      ;  find  n. 


(F,\n~i        /P 
£lj       =  ( L? 


Note.  —  Raise  both  members  of  the  equation  to  the  power  n/(n  —  1). 
19.   In  problem  18  find  n  when  Vi  =  0.5,  Vz  =  6,  PI  =  250  and  P2  = 
7.71. 


CHAPTER  IV 
GRAPHIC  REPRESENTATION 

27.  Modern  life  and  scientific  investigations  make  it  neces- 
sary to  use  statistics  and  to  draw  conclusions  from  statistical 
records.  These  records  usually  consist  of  tabulations  of 
measurements  in  the  form  of  columns  of  figures.  It  is  not 
always  easy  to  get  all  the  information  contained  in  a  table  of 
statistics  without  some  kind  of  pictorial  aid.  When  a  table  of 
numbers  can  be  put  in  a  form  that  appeals  through  the  eye  to 
our  space  notions  it  is  easier  to  understand  and  to  draw  con- 
clusions from  such  a  table. 

Various  representations  to  this  end  are  given  the  name, 
graphic  representations.  Curves,  geometric  diagrams,  pic- 
tures, etc.,  are  frequently  used  for  this  purpose.  Of  the  above 
representations  curves  are  most  widely  used.  In  the  sequel 
our  study  of  graphic  representation  will  be  limited  to  the  method 
of  curves.  The  method  is  best  explained  by  the  use  of  some 
simple  examples. 

1.  Suppose  a  man  walks  for  several  hours  on  a  winding  path 
and  keeps  a  record  of  the  distance  traveled  each  hour.  Let  the 
annexed  diagram  represent  the  path  and  the  points  reached  at 
the  end  of  each  hour. 

1st  hr.,    |  mi.  6th  hr.,  1£  mi. 

2d  hr.,  If  mi.  7th  hr.,  f  mi. 

3rd  hr.,  2  mi.  8th  hr.,  If  mi. 

4th  hr.,  f  mi.  9th  hr.,  \  mi. 

5th  hr.,  \  mi.  10th  hr.,  2  mi. 

To  construct  the  graphic  representation  of  this  record  proceed 
as  follows:  Draw  OT  and  on  it,  beginning  at  0,  lay  off  the  units 

36 


GRAPHS  OF  STATISTICAL  DATA 


37 


of  time  to  any  convenient  scale,  say  \"  to  1  hr.  At  0  draw 
OS  perpendicular  to  OT  and  lay  off  the  units  of  distance  to  a 
scale  of,  say,  \"  to  1  mi.  The  point  0  represents  the  starting 
point,  0  mi.  and  0  hr.  The  point  1  of  the  path  is  represented  at 
li  on  a  A,  1  unit  to  right  of  OS  and  \  unit  above  OT.  The  point 
2  of  the  path  is  represented  at  2i  on  bB,  2  units  to  the  right  of 
OS  and  ^  +  1|  =  2  units  above  OT.  In  a  similar  manner  the 
remaining  points  of  the  path  are  represented  on  the  diagram. 


s 

mi. 

f 

L 

ABCDEFGH 

j 

j 

/ 

10l 

^ 

/ 

ii 

/ 

/ 

f 

^ 

•i. 

/ 

1 

X 

I 

1 

h 

/ 

/ 

-» 

S 

£ 

O    abcdefghtij 

FIG.  7. 


FIG.  8. 


10i  are  now  to  be  connected  by 


The  points  0,  li,  2i,  .  . 
straight  line  segments. 

The  broken  line  Oli,  2i,  .  .  .  10i  in  a  sense  represents  the 
travel  of  the  man  during  a  period  of  10  hrs.  To  continue  the 
study  of  the  diagram  we  need  to  use  certain  terms  which  we 
shall  now  define. 

28.  The  line  OT  is  a  base  line  and  is  called  the  axis  of 
abscissas.  In  this  particular  case  it  is  a  time  axis.  The  line 
OS  is  a  meridian  line  and  is  called  the  axis  of  ordinates.  In  this 


38 


GRAPHIC  REPRESENTATION 


particular  case  it  is  a  distance  axis.  The  distance  of  a  point 
from  the  meridian  is  called  its  abscissa.  The  abscissa  must  be 
measured  to  the  same  scale  as  that  to  which 'the  base  line  is 
laid  off. 

The  distance  of  a  point  from  the  base  line  is  its  ordinate. 
The  ordinate  must  be  measured  to  the  same  scale  as  that  to 
which  the  meridian  is  laid  off. 

The  abscissa  and  ordinate  of  a  point  considered  together  are 
called  its  coordinates.  Evidently  the  position  of  a  point  is 
determined  if  its  coordinates  are  known. 

The  difference  of  the  ordinates  of  two  points  divided  by  the 
difference  of  their  abscissas,  taken  in  the  same  order,  is  called 
the  slope  of  the  straight  line  joining  the  two  points.  The  slope 
is  also  called  the  pitch  of  the  line. 

Resuming  the  study  of  the  example  answer  the  following 
questions:  (a)  What  does  the  ordinate  of  any  of  the  points  li, 
2i,  .  .  .  ,  of  the  broken  line  represent  with  reference  to  the 
travel  of  the  man?  (6)  What  does  the  abscissa  represent? 
(c)  What  does  the  slope  of  any  part  of  the  line  represent  ?  (d) 
Is  the  speed  of  the  man  constant  for  the  whole  time  ?  (e)  What 
is  the  average  speed  of  the  man  for  the  whole  time  ?  Calculate 
the  average  speed  for  the  first  3  hrs.  For  the  second  3  hrs. 
For  the  last  3  hrs. 

2.  A  rod  8"  long  is  set  upright  on  a  level  table  in  sunshine. 
The  lengths  of  the  shadow  were  measured  at  intervals  as 
follows : 


Time  of 

Length  of 

L 

observation  =  t 

shadow  =  I 

j 

f 

10  :  40,  A.M. 

7.79" 

8- 

*i 

^ 

11:15,    " 

7.35" 

' 

11:45,    " 

7.25" 

4- 

12  :  00,  M. 

7.22" 

2- 

12  :  10,  P.M. 

7.22* 

1- 

1    1          1 

1 

,     T 

1  :  00,     " 

7.30" 

0 

s 

o 
o 

'II 

0                   < 

o            < 

§            g 

§        § 

g 

2:15,     " 

7.75" 

o 

IH 

i-l                  C 
IH                  1 

^            fc' 

iH                IM 

CO 

2:30,    " 

9.81" 

FIG.  ( 

J 

GRAPHS  OF  STATISTICAL  DATA  39 

Construct  the  graph  using  time  as  the  abscissa,  but  draw  a 
smooth  curve  instead  of  broken  line.* 

Take  time  scale  \"  to  one  hour,  starting  at  10:00  A.M.  Take 
length  scale  \"  to  1".  Draw  a  second  curve  with  same  time 
scale  but  with  length  scale  \"  to  1".  What  difference  do  you 
notice?  Measure  the  ordinates  at  11:00  A.M.,  and  at  1:30 
P.M.,  and  determine  the  probable  length  of  the  shadow  at  these 
times. 

The  graphic  representation  enables  us  to  determine,  approxi- 
mately at  least,  more  values  than  are  given  in  the  measured 
data.  The  determination  of  new  values  between  the  old  ones 
is  called  interpolation.  Interpolation  is  a  very  useful  applica- 
tion of  the  graphic  representation. 

3.  The  population  of  the  U.  S.  by  decades  was  as  follows, 
in  millions: 


1790    3.9  millions 

1830     12.  8  millions 

1870     38.  5  millions 

1800    5.3        " 

1840     17.0 

it 

1880     50.0 

M 

1810    7.2        " 

1850    23.2 

it 

1890     62.5 

" 

1820    9.7        " 

1860    31.3 

it 

1900     76.3 

1C 

1910    92.0 

(a)  Construct  a  curve  showing  the  probable  population  at 
any  time,  using  tune  as  abscissa. 

(6)  Connect  the  decade  points  of  the  curve  by  straight  lines 
and  determine  the  slope  of  each  segment.  What  information 
does  the  slope  give? 

(c)  Construct  a  curve  on  same  abscissas  as  (a)  using  the 
slopes  found  in  (6)  as  ordinates.  What  information  does  this 
curve  picture? 

4.  The  following  data  were  taken  on  a  certain  machine 
while  in  operation.  The  relative  velocity  of  two  moving  parts 
was  measured  in  ft.  per  sec.  (ft./sec.),  and  the  coefficient  of 
friction  for  each  velocity  was  measured. 

v  =  speed,  ft./sec.    =     135          7        10       15        20 
w  =  coef.  of  friction  =  0.15  0.122  0.104  0.092  0.079  0.066  0.058 

*  The  curve  is  an  attempt  to  represent  the  conditions  between  observed 
points,  the  broken  line  is  not,  except  in  special  cases. 


40  GRAPHIC  REPRESENTATION 

Note.  —  Coefficient  of  friction  is  the  resistance  to  sliding  of 
one  part  on  another  divided  by  the  total  pressure  between  the 
parts.  Construct  a  friction  curve  with  v  as  abscissa. 

What  information  about  speed  and  friction  does  the  curve 
reveal? 

5.  It  costs  $6  to  make  an  article.     The  proprietor  finds  that 
he  can  sell  in  a  given  time  the  following  numbers  of  the  article 
at  the  corresponding  prices: 

Selling  price  =  S  =      $6       $12      $18       $24    $30 
Number  sold  =  N  =  38,000  35,000  24,000  8000  6000 
(a)   Construct  a  sales  curve  using  S  as  abscissa. 
(6)   Calculate  the  profit  at  each  price,  P  =  N  (S  —  6)  and 

construct  curve  with  same  abscissas  as  in  (a)  with  P  as  ordi- 

nates. 

(c)  From  the  sales  curve  estimate  the  number  that  could  be 
sold  at  $15  and  at  $20.     Insert  the  values  of  P  corresponding 
to  these  in  the  profit  curve.     Now  draw  the  profit  curve  so  as 
to  pass  through  these  points. 

(d)  From  the  profit  curve  determine  the  selling  price  that 
will  give  the  greatest  total  profit. 

(e)  What  business  condition  is  indicated  by  the  drop  in  the 
sales  curve  from  $12  to  $24?    What  inference  can  you  draw 
from  the  nearly  equal  sales  at  $24  and  $30?     From  those  at 
$12  and  $6? 

6.  The  magnifying  power  (in  linear  dimensions)  of  a  certain 
spyglass  at  different  distances  from  the  object  viewed  was 
measured  as  follows: 

Magnifying  power  =  m  =  9  8.3  8.1  7.6  7.5  7.6  7.6 
Distance,  meters  =z=23  4  5  6  7  8 
Construct  the  magnification  curve,  using  x  as  abscissa.  What 
general  information  does  the  curve  reveal?  What  do  you 
suspect  regarding  the  correctness  of  the  measurement  7.5? 
The  curve  should  not  be  made  to  pass  through  this  value  as  it 
is  obviously  wrong. 

7.  The  sun's  declination  as  given  in  the  nautical  almanac  for 
1915,  near  the  time  of  the  vernal  equinox,  is  as  follows: 


GRAPHS  OF  STATISTICAL  DATA  41 

March  18  noon  South  1°  05' 

"      19      "  "  0°41' 

"      20      "  "  0°18' 

"      21      "  "  0°06' 

"      22      "  North  0°  30' 

"      23      "  "  0°51' 

Construct  the  declination  curve,  using  time  as  abscissa.  Take 
a  scale  of  TV"  to  1  hr.  This  will  enable  the  determination  of 
hours  with  fair  certainty.  Determine  the  day  and  hour  at 
which  the  curve  crosses  the  axis  of  abscissas.  This  is  the  time 
of  the  vernal  equinox.  What  is  the  time  at  which  the  sun  is 
exactly  over  the  equator? 

Note.  —  The  declination  of  a  heavenly  body  is  its  distance 
north  or  south  of  the  equator  of  the  sky  just  as  latitude  is  the 
distance  of  a  place  on  the  earth  north  or  south  of  the  equator. 
Declination  and  latitude  are  both  measured  in  angular  units, 
that  is,  in  degrees. 

8.  If  s  is  the  amount  (grams)  of  potassium  bromide  that  will 
dissolve  in  100  grams  of  water  at  t°  centigrade,  construct  a 
solubility  curve  from  the  following  data,  using  t  as  abscissa. 

s  =  53.4        64.6        74.6        87.7        93.5 
t  =     0  20  40  60  80 

From  the  curve  determine  the  probable  amount  that  will  dissolve 
at  10°,  30°,  50°,  70°. 

9.  With  the  same  notation  as  in  (8)  the  following  data  were 
taken  for  sodium  nitrate: 

s  =  68.8        72.9        87.5        102 
t  =  -6  0  20  40 

Construct  the  solubility  curve  and  determine  the  probable 
amount  that  will  dissolve  at  10°,  30°. 

10.  Construct  the  temperature  curve  from  the  maximum 
daily  temperatures  at  a  certain  city  for  the  month  of  February. 
For  convenience  assume  the  maximum  occurred  at  the  same 
hour  each  day.    Use  time  as  abscissa. 


42 


GRAPHIC  REPRESENTATION 


Day. 

Temp. 

Day. 

Temp. 

Day. 

Temp. 

Day. 

Temp. 

1 

28 

8 

-2 

15 

10 

22 

26 

2 

36 

9 

14 

16 

13 

23 

13 

3 

31 

10 

4 

17 

21 

24 

20 

4 

14 

11 

1 

18 

24 

25 

26 

5 

24 

12 

11 

19 

11 

26 

35 

6 

28 

13 

15 

20 

20 

27 

42 

7 

18 

14 

14 

21 

30 

28 

42 

11.  A  300-lb.  projectile  is  shot  from  a  10"  rifle  with  a  70-lb. 
charge  of  powder.  The  speed  and  powder  pressure  at  different 
distances  from  the  starting  point  were  measured  as  follows : 


Distance,  ft. 

Measured  speed, 
ft./sec. 

Theoretical  speed, 
ft./sec. 

Pressure, 
Ibs./sq.  in. 

0.41 

436.0 

450.8 

34,752 

0.50 

501.7 

505.0 

34,577 

0.82 

657.0 

661.0 

32,000 

1.06 

750.0 

748.2 

29,648 

1.65 

911.0 

908.0 

24,400 

2.06 

992.8 

992.5 

21,403 

2.26 

1027.0 

1027.0 

19,302 

2.74 

1097.0 

1098.0 

17,580 

3.06 

1137.0 

1139.0 

16,121 

7.06 

1412.0 

1420.0 

7,022 

8.26 

1464.0 

1464.0 

5,476 

10.15 

1527.0 

1516.0 

4,269 

Construct  the  speed  curve  and  the  pressure  curve,  using  dis- 
tance as  abscissa  in  both  cases.  In  fact  both  eurves  may  be 
made  on  the  same  axes  with  different  colored  pencils.  What 
information  regarding  speed  and  pressure  at  different  points 
in  the  gun  do  these  curves  reveal? 

12.  Experiments  with  projectiles  show  that  there  is  a  varia- 
tion of  muzzle  velocity  due  to  a  variation  of  the  temperature 
of  the  powder.  The  curve  on  the  following  page  represents 
the  functional  relation  of  temperature  of  the  powder  and  per 
cent  change  in  muzzle  velocity.  For  temperature  above  70° 
the  arithmetical  correction  of  the  velocity  is  added,  for  tem- 
perature below  70°  it  is  subtracted. 


GRAPHS  OF  STATISTICAL  DATA 


43 


Ti 


44  GRAPHIC  REPRESENTATION 

1.  The  muzzle  velocity  of  a  certain  gun  is  980  ft./sec.  when 
the  temperature  of  the  powder  is  70°.     Show  that  20  ft.  should 
be  subtracted  from  this  velocity  if  the  temperature  of  the 
powder  is  41°. 

2.  The  muzzle  velocity  of  a  gun  is  2250  ft./sec.  when  the 
temperature  of  the  powder  is  70°.     Find  what  correction  should 
be  made  to  this  velocity  when  the  temperature  of  the  powder 
is  71.4°. 

3.  The  muzzle  velocity  of  a  gun  is  2000  ft./sec.  when  the 
temperature   of  the   powder  is   70°.     Find  what   corrections 
should  be  made  to  this  velocity  when  the  temperature  of  the 
powder  is  33.6°. 


CHAPTER  V 
RATIO,  PROPORTION  AND  VARIATION 

29.  Whenever  four  numbers  a,  6,  c,  d  satisfy  the  equation 
(1)  a/6  =  c/d, 

these  numbers  are  said  to  form  a  proportion  or  to  be  in  propor- 
tion. The  equation  (1)  contains  two  equal  fractions  or  ratios. 
When  we  measure  any  quantity  by  comparing  it  with  another 
of  the  same  kind  we  are  said  to  measure  the  first  quantity  by 
the  second  as  a  unit.  The  number  which  expresses  how  many 
units  and  parts  of  units  result  from  the  comparison  is  the 
measure  of  the  quantity  to  the  unit  employed. 

30.  A  ratio  is  defined  as  the  measure  of  any  quantity,  ab- 
stract or  concrete,  when  some  other  quantity  of  the  same  kind 
is  the  unit  of  measure.     When  the  length  of  a  room  is  measured 
with  a  foot  rule  (one-foot  length)  and  there  is  obtained  24  as 
the  measure  or  numerical  value  of  the  length  of  the  room  to 
that  unit,  it  is  said  the  length  of  the  room  is  24'.     This  implies 
,^\  length  of  room  _  _ . 

one  foot  length 

This  is  the  idea  involved  in  all  measurements  where  definite 
units  are  available. 

If  the  length  of  the  room  is  measured  with  a  yard  stick  (one 
yard  length),  the  measure  of  the  room  is  8.     The  room  is  said 
to  be  8  yds.  long.    This  implies 
,„*  length  of  room  _  _ 

one  yd.  length 

A  ratio  is  to  be  regarded  as  a  mere  number  without  material 
denomination,  that  is,  it  is  an  abstract  number.  Since  a  ratio 
is  essentially  a  quotient  or  an  indicated  division,  it  is  con- 

45 


46  RATIO,  PROPORTION  AND  VARIATION 

veniently  represented  by  a  fraction.  A  ratio  is,  therefore, 
subject  to  all  the  arithmetical  operations  ordinarily  performed 
with  fractions. 

31.  From  (1)  and  some  general  axioms  regarding  equalities 
it  is  easy  to  establish  the  following  fundamental  and  useful 
theorems: 

a       c 
If  T  =  -,,  then  it  can  be  proved  that: 

1.  ad  =  be,  the  product  of  the  extremes  equals  the  product 
of  the  means. 

2.  -  =  -j  ,  proportion  by  alternation. 

C        CL 

3.  -  =  -  ,  proportion  by  inversion. 

Cl         C 


5. 
6. 
7. 

b 

a  — 

6 

c 

d 

i 

d 

proportion  by  division. 

proportion  by  composition  and  division. 
,       a+c+e+0           a     c     e 

g 

b 

a  + 

b. 

c 

d 

'  i 

d 

Tf  a 
6 

b 

c 
c 

d~ 

e  _ 

d' 
0 

h 

+d+f+h 

~b 

d 

f 

h 

32.  Variation.  —  The  statements,  "  x  is  proportional  to  y," 
"x  varies  as  y,"  "x  is  directly  proportional  to  y,"  "x  varies 
directly  as  y,"  are  all  different  ways  of  saying  the  same  thing. 
This  relation  between  x  and  y  may  be  put  in  the  form 

(1)  .    x  =  ky,  * 
where  A;  is  a  fixed  value,  and  x,  y  take  any  number  of  different 
values  which  satisfy  this  equation. 

The  statements,  ('x  varies  inversely  as  y,"  "x  is  inversely 
proportional  to  y,  "  are  each  equivalent  to  the  equation 

k 

(2)  xy  =  k  or  x  =  -, 

y 

where  k  is  constant  as  in  equation  (1).  J 


VARIATION  47 

The  statements,  "x  is  jointly  proportional  to  y  and  z"  "x 
varies  jointly  as  y  and  z,"  are  each  equivalent  to  the  equation 

(3)  x  =  kyz. 

From  type  forms  (1),  (2)  there  follows  a  single  equation 

(4)  ,-*• 

Equations  (1),  (2),  (3),  (4)  are  to  be  regarded  as  special  and 
convenient  working  forms  of  proportion.  Many  laws  of  nature 
are  expressed  in  one  or  another  of  these  forms. 

Illustrations.  —  /  =  k     „    ,  the  law  of  gravitation;  A  =  xy, 

the  area  of  a  rectangle;  E  =  kn2,  the  law  of  mob  energy. 

1.  Suppose  it  is  known  that  s  varies  as  t,  and  that  s  =  25, 
when  t  =  5.     To  find  the  equation  of  relation  between  s  and  t 
and  to  find  the  value  of  s  when  t  =  10. 

Write  by  (1), 

s  =  kt. 

Substituting  values  given  in  the  problem, 

25  =  k  -  5, 

therefore  k  =  5. 

The  desired  equation  is 

s  =  5t. 
The  value  of  s  when  t  =  10  is 

s  =  5 . 10  =  50. 

2.  The  carrying  capacity  of  pipes  varies  as  the  squares  of 
their  diameters  (friction  neglected).     How  many  6"  pipes  will 
carry  as  much  as  one  24"  pipe  ? 

Call  the  carrying  capacity  of  a  pipe  Cf  and  write 

(1)  C  =  kd* 

for  any  and  all  pipes.     For  6"  pipes  then, 

(2)  Ce=?  fc-62 


48  RATIO,  PROPORTION  AND  VARIATION 

(3)  Also  C24  =  k  -  24?  =  576  k. 
By  equations  (2)  and  (3) , 

Cu      576 

(4)  -^j-  =  -^-  =  16,  the  required  number  of  6"  pipes. 
0$       06 

Note.  —  In  this  case  we  did  not  determine  the  value  of  k. 
Its  presence  in  the  equations  gave  us  all  the  advantage  of 
knowing  its  value  without  finding  it. 

3.  The  weight  of  a  sphere  of  given  material  varies  as  the  cube 
of  its  radius.     If  a  sphere  of  1'  radius  weighs  600  Ibs.,  what 
will  a  sphere  of  radius  5^'  of  same  material  weigh? 

Note.  —  Solve  this  and  the  following  by  the  methods  of  Ex. 
(1),  (2),  above. 

4.  The  capacity  of  a  cylindrical  tank  varies  as  its  height 
when  the  diameter  is  fixed  and  as  the  square  of  the  diameter 
when  the  height  is  fixed.     A  tank  1'  in  diameter  and  1'  high  has 
a  capacity  of  5.83  gallons.     Find  the  capacity  of  a  tank  8'  in 
diameter  and  30'  high. 

Note.  —  Write  C  =  k> &- h. 

5.  The  value  of  a  diamond  varies  as  the  square  of  its  weight. 
A  diamond  worth  $600  is  cut  in  two  pieces  whose  weights  are  as 
1  to  3.     What  is  the  value  of  each  piece? 

6.  The  illumination  from  a  light  varies  inversely  as  the 
square  of  the  distance  from  the  light.     If  an  object  10"  from 
the  light  be  moved  10  (Vo  —  1)"  farther  away,  what  is  the 
ratio  of  the  final  to  the  original  illumination.     (Assume  I  as 
the  original  illumination.) 

7.  The  cross  section  of  a  chimney  should  vary  as  the  quan- 
tity of  fuel  used  per  hr.  and  inversely  as  the  square  root  of  the 
height.     The  cross  section  of  a  chimney  150'  high  is  30  sq.  ft., 
the  quantity  of  fuel  used  per  hr.  is  15,000  Ibs.     Find  the  cross 
section  of  a  chimney  100'  high  connected  to  a  furnace  using 
5000  Ibs.  of  fuel  per  hr. 

8.  A  solid  spherical  mass  of  clay  4"  in  diameter  is  moulded 
into  a  spherical  shell  whose  outside  diameter  is  6".    What  is 


SUPPLEMENTARY  EXERCISES  49 

the  inside  diameter  of  the  shell?    It  is  given  that  the  volume  of 
a  sphere  varies  as  the  cube  of  its  diameter. 

9.  A  safe  load  on  a  horizontal  beam  supported  at  its  ends 
varies  as  the  breadth  and  as  the  square  of  the  depth  and  in- 
versely as  the  length  of  the  beam.    A  beam  2"  x  6"  x  12',  on 
edge,  will  sustain  a  load  of  700  Ibs.    What  load  will  a  beam  of 
the  same  material  3"  x  9"  x  18',  on  edge,  sustain? 

10.  The  weight  of  a  body  above  the  earth's  surface  varies 
inversely  as  the  square  of  its  distance  from  the  center  of  the 
earth.     If  a  body  weighs  150  Ibs.  just  outside  the  surface,  how 
high  must  it  be  raised  so  its  weight  will  be  30  Ibs.,  the  radius 
of  the  earth  being  assumed  to  be  4000  mi? 

11.  A  tree  casts  a  shadow  70'  long  on  level  ground.     At  the 
same  time  a  10'  pole  casts  a  shadow  9'  long.     Find  the  height 
of  the  tree. 

12.  The  velocity  of  a  falling  body  starting  from  rest  varies 
as  the  time.    At  the  end  of  2  sec.  the  velocity  is  64.32'  per  sec. 
What  is  the  formula  holding  between  velocity  and  time  and 
what  is  the  velocity  at  the  end  of  4  seconds? 

Note.  —  Assume  v  =  k  •  t. 

13.  The  interest  on  a  given  principle  varies  jointly  as  the 
time  and  rate.     $500  yields  $25  in  two  years.     What  is  the  rate  ? 

14.  If  the  amount  of  fuel  required  to  heat  a  house  varies  as 
the  square  of  the  difference  in  temperature  in  the  house  and 
outside  and  if  when  the  thermometer  reads  0°  C  outside,  and 
the  temperature  inside  is  20°  C,  it  requires  1000  cu.  ft.  of  gas 
per  hr.  to  heat  a  certain  house,  how  much  gas  would  be  neces- 
sary to  heat  the  house  to  the  same  temperature  when  the  ther- 
mometer outside  dropped  to  — 10°  C  ? 

SUPPLEMENTARY  EXERCISES 

1.  If  y  varies  as  x  and  if  x  =  4  when  y  =  f ,  find  y  when  x  =  3. 

2.  If  y  is  proportional  to  x  and  if  x  =  £  when  y  =  f ,  find  x  when  y  =  7. 

3.  If  y  varies  directly  as  x  and  inversely  as  z  and  if  y  =  8  when  x  =  12 
and  2  =  f ,  find  z  when  y  =  2  and  x  =  6. 

4.  If  y  varies  jointly  as  x  and  z  and  if  y  =  f  when  x  =  5  and  z  =  f , 
find  z  when  y  —  3  and  x  =  6. 


50  RATIO,   PROPORTION  AND  VARIATION 

5.  If  y  varies  as  x  and  x  =  6  when  y  =  54,  what  is  the  value  of  y  when 
x  =  8? 

6.  If  x  varies  directly  as  y  and  inversely  as  z,  and  x  =  f  when  y  =  20 
and  z  =  10,  what  is  the  value  of  x  when  y  —  9  and  2  =  20  ? 

7.  If  x4  is  directly  proportional  to  y3  and  x  =  4  when  y  —  4,  what  is  the 
value  of  x  when  y  =  9  ? 

8.  If  2  x  —  3  is  proportional  to  y  +  6  and  x  =  4  when  y  =  3,  what  is 
the  value  of  y  when  x  =  5? 

9.  The  hypotenuse  of  a  right  triangle  is  100  ft.  long.     Find  the  other 
sides,  if  their  ratio  is  3  to  4. 

10.  The  stretch  in  the  spring  of  an  ordinary  spring  balance  is  propor- 
tional to  the  weight  (force)  applied.     If  a  force  of  4  Ibs.  stretches  it  one 
inch,  how  much  will  a  force  of  17  Ibs.  stretch  it? 

11.  How  would  you  graduate  (divide)  a  scale  for  the  spring  balance  in 
the  above  problem? 

12.  The  area  of  a  circle  varies  as  the  square  of  its  diameter.     How  is 
the  diameter  affected  when  the  area  is  doubled?     How  is  the  area  affected 
when  the  diameter  is  doubled?     Give  work  showing  reasons  for  your 
answers. 

13.  If  two  pulleys  are  connected  with  a  belt  prove: 

(a)   That  the  number  of  revolutions  that  they  make  per  minute  are  to 
each  other  inversely  as  their  diameters. 

(6)   That  their  radii  are  to  each  other  inversely  as  their  speeds. 

14.  If  x  varies  as  y,  then  x  =  ky  and  conversely,  if  x  =  ky,  then  x  varies 
as  y.     Show  that  the  speed  of  a  point  on  the  rim  of  a  pulley  varies  as  its 
diameter. 

15.  If  in  problem  13  the  driving  pulley  is  making  20  revolutions  per  min- 
ute, and  its  diameter  is  10  inches,  find  the  number  of  revolutions  per 
minute  of  the  second  pulley  if  it  is  4  inches  in  diameter. 

16.  According  to  Boyle's  law  of  gases,  pressure  (p)  times  volume  (v)  is 
constant.     How  does  the  pressure  vary  with  the  volume?     Show  graphi- 
cally the  relation  between  (p)  and  (v)  if  v  =  1  cu.  ft.  when  p  =  25  Ibs.  per 
sq.  in. 

17.  Given  that: 


L  =  2-xrh  and  L'  = 

A=2irr(r  +  h]  and  A'  =  2^'  (r'  +  fc'), 

formulas  which  represent  the  total  area  and  lateral  area  of  two  right 
circular  cylinders.     Show  that 

IL  -  A  =  z!  =  ^1 

L'      A'      r'*      h1*' 
18.   In  reading  contour  maps  the  question  arises  whether  a  station  B  is 


SUPPLEMENTARY   EXERCISES 


51 


visible  from  some  station  A.     The  problem  is  to  determine  whether  an 
intervening  height  of  ground  C  obstructs  the  line  of  sight  from  A  to  B. 


FIG.  10. 


The  horizontal  distance  between  A  and  B  as  scaled  on  the  map  is  1500 
yds.,  between  A  and  C  is  900  yds.,  and  height  of  B  above  datum  plane 
(horizontal  plane  through  station  A)  is  80  ft.,  height  of  obstacle  C  above 
datum  plane  is  50  ft.  Determine  whether  A  is  visible  from  B.  By  means 
of  the  figure  and  theorem  27,  Chapter  II,  it  is  seen  that: 


and  solving 


900  : 1500  =  h  :  80 

h  =  48. 


Therefore,  the  line  joining  stations  A  and  B  would  pass  station  C  at  a 
height  of  48  ft.  above  the  datum  plane.  Since  obstacle  C  is  50  ft.  above 
this  plane  it  obstructs  the  vision  from  B  to  A. 

19.  Check  the  above  problem  by  drawing  the  figure  to  scale. 

20.  In  problem  18  substitute  y  for  900,  x  for  1500,  H  for  80,  and  solve 
the  equation  for  h.     Draw  a  figure  and  indicate  the  distances  x,  y,  H  and  h. 

21.  Using  the  result  of  problem  20,  find  whether  C  would  obstruct  the 
line  of  sight  from  B  to  A  if  x  =  2100  yds.,  y  =  1400  yds.,  H  =  130  ft.  and 
the  height  of  C  above  datum  plane  is  95  ft. 


CHAPTER  VI 


THE  RECTANGULAR  COORDINATE   SYSTEM: 
GRAPHS   OF  EQUATIONS;  FORMULAS 

33.  Two  lines  are  selected,  intersecting  at  right  angles,  in 
the  plane  of  the  paper  (they  are  usually  chosen  the  one  hori- 
zontal and  the  other  vertical).  The  position  of  a  point  is 
known  if  its  distances  from  these  two  lines  are  known.  Let 


FIG.  11. 

XX'  and  YY'  intersect  at  right  angles  in  0.  The  position  of 
P  is  known  if  Om  =  x  and  mP  =  y  are  known.  The  point  P 
will  often  be  designated  by  P  (x,  y)  or  by  (x,  y),  at  pleasure. 
It  is  noted  that  the  abscissa  x  is  always  written  before  the 
ordinate  y  and  this  arrangement  holds  no  matter  what  letters 
are  used  to  represent  the  coordinates. 

52 


GRAPHS  OF  FUNCTIONS  AND  EQUATIONS     53 

Since  the  point  P  may  lie  either  to  the  right  or  left  of  YY' 
and  above  or  below  XX',  we  must  have  further  conventions. 
The  following  have  been  universally  adopted: 

The  ordinates  of  points  above  the  XX'  axis  are  positive. 

The  ordinates  of  points  below  the  XX'  axis  are  negative. 

The  abscissas  of  points  to  the  right  of  YY'  axis  are  positive. 

The  abscissas  of  points  to  the  left  of  YY'  axis  are  negative. 

The  point  0  is  called  the  origin  of  coordinates  or,  for  short, 
the  origin.  Both  its  coordinates  are  zero. 

1.  Locate  the  following  points  in  rectangular  coordinates: 
(2,  -5);   (-4,  1);   (-3,  -3);  (4,5);   (1,8). 

2.  What  is  the  abscissa  of  any  point  on  YY't    What  is  the 
ordinate  of  any  point  on  XX'  ? 

3.  What  is  the  ordinate  of  any  point  of  a  line  parallel  to  XX' 
and  6  units  above  it  ?  -  On  what  line  do  all  points  whose  abscissas 
equal  —  3  lie? 

34.  Graphs  of  functions  *  and  equations.  —  To  construct 
the  graph  of  any  function,  say  3  z2  —  4  x  +  5,  write  it  equal  to 
some  symbol,  say  y,  obtaining  an  equation, 

(1)  2/  =  3z2-4z  +  5. 

Select  one  of  the  symbols  x  and  y  for  the  abscissa  and  the  other 
as  the  ordinate  of  points  on  the  graph.  In  this  and  similar 
cases,  x  is  usually  the  abscissa  and  y  the  ordinate. 

Every  equation  like  (1)  expresses  a  relation  between  different 
number  pairs.  Every  number  pair  are  the  coordinates  of  a 
point.  One  number  of  each  pair  may  be  chosen  at  pleasure, 
the  other  number  of  the  pair  must  be  calculated  from  the 
equation.  Any  number  of  such  number  pairs  may  thus  be 
determined.  The  corresponding  points  can  be  located  on  the 
diagram.  Having  located  several  points,  draw  a  smooth  curve 
through  these  points  as  in  Chapter  IV.  This  curve  is  called 
the  graph  or  the  locus  of  the  equation. 

It  must  be  remembered  that  the  coordinates  of  all  points  on 

*  For  definition  of  function  see  Chapter  VII.  The  term  formula  might 
be  used  at  present. 


54 


the  graph  must  satisfy  the  equation.  On  the  other  hand  every 
number  pair  which  satisfy  the  equation  must  be  coordinates  of 
a  point  on  the  graph.  To  be  sure,  the  absolutely  exact  graph 
cannot  be  drawn  without  determining  the  coordinates  of  every 
point  on  it.  This,  obviously,  would  be  impossible  for  it  would 
involve  endless  calculation  to  obtain  even  a  small  portion  of  the 
graph.  But  by  carefully  determining  a  few  well-selected  points 
of  the  graph  and  then  drawing  a  smooth  curve  through  these 
points,  it  will  be  found  that  a  very  close  approximation  to  the 
true  graph  is  obtained.  This  approximate  graph  is  sufficiently 
accurate  to  permit  interpolation  and  to  furnish  a  good  basis  for 
a  study  of  the  equation  it  represents  and  of  the  true  graph. 

The  process  of  con- 
structing the  graph  of 
equation  (1)  is  indicated 
below.  Choose  values  of 
x  and  calculate  the  cor- 
responding values  of  y 
from  the  equation, 

x=     -2,       -1,     0, 

+  1,     +2,     +3 
y  =  +25,     +12,    +5, 
+4,     +9,     +20 

The  annexed  figure  shows 
the  approximate  graph 
•X  which  gives  a  good  idea 
of  the  true  graph.  Owing 
to  the  large  values  of  y, 
a  smaller  scale  is  used 
for  ordinates  than  for 
abscissas  in  order  to  reduce  the  space  necessary  to  draw  a 
sufficient  portion  of  the  curve.  This  distortion  modifies  the 
form  of  the  curve  but  does  not  affect  its  fundamental  nature. 
It  reduces  the  distance  between  the  calculated  points  of  the 
curve,  see  17. 


FIG.  12. 


GRAPHS  OF  FUNCTIONS  AND  EQUATIONS 


55 


2.   Construct   the   graph   of   the   equation   2  x  +  3  y  =  5. 
Assign  values  to  x  and  calculate  values  of  y  from  the  equation. 
x=  -3    2        5          8 

y    =          gl  1  ^2  JJ2 

This  locus  is  a  straight  line.  What  is  its  slope?  At  what  value 
of  x  does  the  line  cross  the  axis  XX"!  This  value  is  the  x- 
intercept  of  the  line.  At  what  value  of  y  does  the  line  cross  the 
axis  YY'f  This  value  is  the  ^-intercept  of  the  line. 

Y 


FIG.  13. 

A  straight  line  is  always  associated  with  an  equation  of  the 
first  degree  in  two  unknowns. 

3.  Construct  the  graph  of  the  equation  x2  +  y2  =  25.  As 
in  the  preceding  cases  choose  values  of  x  and  calculate  the 
corresponding  values  of  y  from  the  equation. 

x=  -6         -5         -4     -3     -2         -1 
y=    (i)*          0        ±3     ±4     ±4.6     ±4.9 
x  =  +1         +2        +3     +4         +5     +6 
y=±4.9     ±4.6     ±4     ±3  0        (i)* 

It  is  noted  that  to  every  value  of  x  there  correspond  two  equal 

but  oppositely  signed  values  of  y.     The  curve  is,  therefore, 

*  (i)  represents  an  imaginary  quantity. 


0 

±5 


56 


THE  RECTANGULAR  COORDINATE  SYSTEM 


FIG.  14. 


symmetrical  about  the  axis  XX'.  If  values  of  y  are  chosen  and 
values  of  x  calculated  from  the  equation  it  will  be  found  that 
to  every  value  assigned  to  y  there  cor- 
respond two  equal  and  opposite  values 
of  x.  Therefore  the  curve  is  symmet- 
rical about  the  axis  YY'.  Conse- 
quently the  curve  is  symmetrical  about 
the  origin.  This  curve  is  a  circle.  Its 
radius  is  5  and  its  center  is  at  0.  A 
circle  is  always  associated  with  an 
equation  of  the  form  x2  +  yz  =  a2, 
that  is,  the  sum  of  two  squares  equals 
a  constant. 

What  is  the  nature  of  the  values  of  y  for  all  values  of  x  <  —  5  ? 

y  "         x  >      5? 

x  y  <  -5? 

"  "  "          x  "         y  >      5? 

It  is  seen  that  no  real  values  of  x  or  y  outside  the  limits  —5  and 
+5  will  satisfy  the  equation.  This  means  that  no  points  of 
the  curve  can  lie  outside  these  limits. 

4.  The  corresponding  values  of  two  symbols  x  and  y  are  such 
that  the  square  of  one  equals  the  square  of  the  other  increased 
by  1.    Write  the  equation  and  construct  the  graph.    Save 
graph  for  Ex.  7. 

5.  The  corresponding  values  of  two  symbols  are  such  that 
the  sum  of  their  squares  equals  49.     Write  the  equation  and 
construct  the  graph.     Save  graph  for  Ex.  7. 

6.  Construct  the  graphs  of  y  =  2  x  and  of  y  =  2  x  +  6  on 
the  same  axes  and  to  same  scale.     What  likeness  and  what 
difference  do  you  notice  regarding  the  lines?    What  likeness 
and  what  difference  regarding  the  equations? 

What  is  the  slope  of  each  line  ?  What  is  the  coefficient  of  x 
in  each  equation?  Can  you  tell  the  slope  of  a  straight  line 
from  its  equation  without  drawing  the  line?  How? 

7.  Determine  intercepts  on  the  axes  of  all  the  lines  in  the 
last  three  exercises. 


EMPIRICAL  EQUATIONS  57 

8.  Construct  the  graph  of  x  +  4  y  =  14  and  determine  the 
slope  and  the  intercept  on  the  axes. 

9o  Construct  the  graphs  of  rc/3  +  y/5  =  9  and  8  x  *-  4  y  + 
4  =  0  on  the  same  diagram.  Determine  by  measurement  the 
coordinates  of  the  point  of  intersection  of  these  graphs.  Solve 
the  given  equations  as  simultaneous  equations  for  x  and  y. 
Compare  the  results  with  the  coordinates  of  the  point  of  inter- 
section. Explain. 

10.  Treat  the  equations  x2  +  yz  =  25  and  x  —  y  =  1,  in  a 
manner  similar  to  that  directed  in  Ex.  9. 

11.  Treat  the  equations  x  +  y  =  6  and  x  —  y  =  1   in  a 
manner  similar  to  that  directed  in  Ex.  9. 

12.  Draw  the  triangle  whose  sides  lie  in  the  lines  represented 
by  the  equations:  x  +  y  =  4;  2x  +  y  =  —2;  x  —  y  =  —6. 

Note.  —  The  vertices  of  the  triangle  will  be  the  three  points 
of  intersection  of  the  lines  in  pairs. 

13.  A  sphere  of  wood  1'  in  diameter  sinks  in  water  to  a  depth 
determined  by  one  root  of  the  equation  2  x3  —  3  x2  +  0.657  =  0. 

Note.  —  The  desired  root  will  be  one  of  the  x-intercepts  of 
the  curve,  i/  =  2z3-3z2  +  0.657.  Explain. 

35.  An  important  use  of  graphs  is  found  in  the  determina- 
tion of  empirical  formulas  expressing  laws  of  nature.  From 
observations  made  in  the  laboratory  or  in  the  field  several 
number  pairs  are  measured.  From  these  a  curve  is  constructed. 
This  curve  represents  to  the  eye  the  relation  between  the 
numbers  of  each  of  the  number  pairs.  The  form  of  the  curve 
may  often  suggest  to  the  experienced  mathematician  the 
general  form  of  an  equation  of  which  the  curve  is  the  graph. 
It  remains  to  determine  the  constants  of  this  equation.  Some- 
times only  an  approximately  correct  formula  can  be  determined 
at  first.  This  formula  is  subject  to  later  correction  by  addi- 
tional observations  and  by  the  application  of  least  squares. 

A  few  examples  will  illustrate  the  meihod  of  procedure. 

1.  Let  us  attempt  to  find  an  equation  for  problem  6,  28, 
Chapter  IV.  The  form  of  the  curve  suggests  to  one  expe- 
rienced in  the  art,  an  equation  of  the  form  xy  ~  k  or  y  = 


58  THE  RECTANGULAR  COORDINATE  SYSTEM 

-  +  6  as  possibilities  where  either  k  or  else  a  and  b  are  to  be 
determined.  Let  us  try  the  second  form  and  write 

(1)  m  =  l  +  b. 

There  being  two  unknown  constants  we  shall  need  two  equa- 
tions. These  are  obtained  by  substituting  in  (1)  two  pairs  of 
observed  values  of  m  and  x.  Thus 

(2)  9  =  !  +  &> 

(3)  7.6  =  ^  +  6. 

o 

Solving  these  equations  simultaneously  for  a  and  6  gives 
a  =  4.7  and  6  =  6.7.  Substituting  these  in  equation  (1)  gives 

as  the  desired  formula: 

« 

(4)  m  =  ^  +  6.7. 

To  check  the  validity  of  this  formula  for  values  not  included  in 
determining  a  and  b,  put  x  =  4  and  m  turns  out  to  be  7.9. 
The  corresponding  observed  value  is  8.1.  The  formula  gives 
fairly  good  results  considering  the  nature  of  the  quantities 
concerned. 

2.  An  innkeeper  finds  that  if  he  has  G  guests  per  day  his 
expenses  are  $E  and  his  receipts  $R.  His  books  furnish  the 
following  data: 

G  =  210  270  320  360 
#=83  97  108  117 
R=  79  106  132  149 

Using  G  as  abscissa  construct  two  graphs,  one  for  E  and  one  for 
R  on  same  axes  and  same  scale.  These  lines  appear  to  be 
straight  lines.  This  fact  suggests  an  equation  for  each  of  the 
form  y  =  mx  +  6.  Write 


EMPIRICAL  EQUATIONS 


59 


(1) 

(2) 


E  = 
R  = 


+  bi. 

+  62. 


The  values  of  m\,  mz,  &i,  62  may  be  determined  easily  from  the 


graph.    The  slopes  are  mi, 
d'c' 


dc 
Measure  mi  =  ji  =  0.203; 


mz  =  jTr,  =  0.444.    The  ^/-intercepts  are  ob  =  bi  =  32;   ob'  = 
bz  =  —35.     These  values  are  to  be  regarded  as  close  approxi- 


150- 


-Id'. 


FIG.  15. 


mations,  only.     The  equations  (1)  and  (2)  now  become,  by 

substituting  these  values, 

(!')  E  =  0.203  G  +  32. 

(2')  R  =  0.444  G-  35. 

(a)   What  number  of  guests  is  just  sufficient  to  pay  expenses  ? 

(6)  What  are  the  expenses  when  G  is  zero?  What  are  the 
profits  when  G  =  360?  How  can  profits  and  losses  be  deter- 
mined from  the  graph? 


60  THE  RECTANGULAR  COORDINATE  SYSTEM 

Equations  (!')  and  (2')  can  be  obtained  algebraically.  Sub- 
stitute two  pairs  of  values  of  E  and  G  in  (1)  and  solve  for 
mi,  61.  Then  substitute  two  pairs  of  values  of  R  and  G  in  (2) 
and  solve  for  mz  and  62.  Thus  from  (1) 

83  =  Wi  •  0.210  +  61, 
117  =  mi-  360  +  6!. 

Whence  mi  =  0.233  and  61  =  35.  In  a  similar  manner  with 
Eq.  (2): 

79  =  ma  •  210  +  62, 
149  =  mz  •  360  +  62. 

Whence  mz  =  0.466  and  62  =  29.  These  values  are  in  fair 
agreement  with  the  ones  determined  above. 

3.  The  latent  heat  of  steam  at  6°  C.  is  L.     Construct  a  curve 
from  the  values  given  below  and  determine  an  equation  of  the 
form  L  =  md  +  b. 

4.  The  height  h,  above  the  earth's  surface  and  the  corre- 
sponding barometer  reading  p,  are  as  follows: 

h  (ft.)  =    0        886        2703        4763        6942 
p  (in.)  =  30          29  27  25  23 

Construct  the  graph  and  determine  an  equation  of  the  form 

p  =  Aekh. 

Note.  —  Take  the  logarithm  of  both  sides  of  the  proposed 
equation  and  proceed  as  in  previous  cases.  Thus 

log  p  =  log  A  +  kh  log  e,      e  =  2.718. 

Determine  first,  log  A  and  k  as  in  previous  cases.  The  value 
of  A  and  k  are  then  to  be  substituted  in  the  proposed  equation 
p  =  A&h. 

5.  Determine  the  equation  of  the  form  y  =  ax2  +  bx  +  c 
of  the  curve  which  passes  through  the  points  given  by 

x=    2  3  4          5 

y  =  10        -6        -5        10 


EMPIRICAL  EQUATIONS  61 

6.  Construct  the  curve  and  determine  the  equation  of  the 
fox-m  y  =  y?  +  bx*  -+-  ex  +  d  from  the  following  data: 

x=  -2        -1        0        1        2          3 
y  =  -8        -1        0        1        2        27 

7.  Construct  the  graph  and  determine  an  equation  of  the 
form  yz  =  ax  +  6  from 

x  =  2         4 
y  =  4        10 

Determine  whether  or  not  the  points  (0,  0),  (3,  7)  are  off  the 
curve. 

8.  Determine  an  equation  and  draw  the  curve  from 

x  =  Q        136 
t/  =  6        5.9        5.3        0 

The  equation  is  to  be  of  the  form  ax2  +  by2  =  c2. 

9.  For  an  ideal  gas  Boyle's  law  says  that  the  product  of  the 
pressure  and  volume  of  a  given  mass  of  gas  at  constant  tem- 
perature is  constant.     Form  an  equation  from  this  statement 
and  draw  the  graph.     Determine  an  equation  from  the  follow- 
ing: 

Pressure,  inches  of  mercury  =  130  45      60  75  90      105 
Volume,  cubic  centimeter     =  100  66.6  50  40  33.3  28.5 

10.  The  intensity  of  illumination  from  a  light  varies  inversely 
as  the  square  of  the  distance  from  the  light.     Write  this  in  the 
form  of  an  equation  and   draw  the  graph.     Determine  an 
equation  from 

L  =  100   50      25    5 
Z>=  10    14.14    20   44 

Note.  —  If  desired  the  subject  of  empirical  formulas  may  be 
continued  at  this  time  by  taking  up  the  work,  which  appears 
in  a  later  chapter,  on  the  applications  of  logarithmic  and  semi- 
logarithmic  paper  in  determining  certain  types  of  empirical 
formulas. 


CHAPTER  VII 
NUMBERS,  VARIABLES,  FUNCTIONS,  LIMITS 

36.  Numbers.  —  (a)  The  numbers  1,  2,  3,  ...  used  in 
ordinary  counting  are  called  the  natural  or  absolute  integers. 
The  idea  of  positive  and  negative  does  not  belong  to  them. 
They  answer  the  question,  "How  many?"  We  associate  with 
the  natural  integers  all  fractions  formed  with  them,  such  as,  |, 
f,  -VS  ....  The  natural  integers  and  the  associated  fractions 
constitute  the  number  system  of  ordinary  arithmetic.  They 
are  sometimes  called  non-directed  numbers.  They  are  some- 
times, but  with  questionable  propriety,  identified  with  the 
positive  numbers  of  algebra. 

(6)  The  idea  of  positive  and  negative  numbers  is  a  relatively 
modern  notion.  Certain  natural  phenomena,  scientific  meas- 
urements and  the  fact  that  subtraction  was  not  always  possible 
with  the  natural  numbers  suggested  the  idea  of  "opposite" 
numbers  or  positive  and  negative  numbers. 

(c)  A  rational  number  is  one  that  can  be  expressed  as  the 
quotient  of  two  integers,  positive  or  negative.     Some  rational 
numbers  are  expressible  in  decimal  form  as  f,"|  =  0.4,  —  |  = 
—0.125,  etc.     Some  rational  numbers  cannot  be  exactly  ex- 
pressed in  decimal  form,  as  f  =  0.666  .  .  .  ,  |  =  0.1111  .... 

(d)  Irrational  numbers  jesult  from  certain  operations  on 
rational  numbers.     Thus  \/2  =  1.4142  ...  is  irrational.     For 
it  cannot  be  expressed  as  the  quotient  of  two'  integers.    The 
numbers  V3,  v^,  TT  =  3.14159  .  .  .  are  examples  of  irrational 
numbers.     The  logarithms  of  most  numbers  are  irrational. 

(e)  All  the  numbers  so  far  mentioned  come  under  a  more 
general  class  called  real  numbers.     The  name  is  derived  from 
their  association  with  ordinary  affairs  and  by  contrast  with  a 
class  of  numbers  defined  below. 

62 


A   VARIABLE  63 

(/)  An  imaginary  number  is  one  that  arises  in  the  attempt 
to  take  an  even  root  of  a  negative  real  number.  Thus  \/— 2, 
v'—  3,  .  .  .  are  imaginary  numbers. 

Remark.  —  It  should  be  noted  that  the  name  "imaginary' 
applied  to  numbers  reflects  an  attitude  of  mind,  existing  for- 
merly, toward  such  numbers.  Recent  developments  have 
shown  that  term  is  unfortunate.  For  imaginary  numbers  have 
come  to  have  a  very  real  meaning  in  scientific  investigations. 
Attention  seems  first  to  have  been  directed  to  imaginary 
numbers  in  the  study  of  quadratic  equations. 

(gr)  The  sum  of  a  real  number  and  an  imaginary  number  is 
called  a  complex  number.  All  complex  numbers  are  of  the  form 
a  ±  6  V^I  where  a,  6  are  real.  Thus  2  +  V^9  =  2  + 
3  V^;  -|  +  |  V^3  =  -|  +  £  V3  •  V^l  are  complex 
numbers.  Use  will  be  made  of  complex  numbers  in  a  later 
chapter. 

It  is  to  be  kept  in  mind  that  in  practical  calculations  involving 
irrational  numbers  it  is  necessary  to  use  a  rational  number 
nearly  equal  to  the  corresponding  irrational  number.  Thus  for 
\/2  the  rational  number  1.4142,  which  is  correct  to  five  figures, 
may  be  used,  and  similarly  in  other  cases. 

37.  (a)  A  variable  is  a  symbol  to  which,  in  a  given  problem, 
may  be  assigned  an  indefinitely  great  number  of  values,  and 
which  may  be  employed  in  calculations  as  a  number.  The 
values  assigned  to  a  variable  may  be  assigned  in  accordance 
with  some  law  of  nature  or  in  accordance  with  some  arbitrary 
mode  of  thought. 

May  x  and  y  have  more  than  one  value  each  in  the  equation 
y  —  2  x  =  4?  Does  2  ever  have  any  value  different  from  2? 
Are  x  and  y  variables  or  fixed  in  the  equation  ?  Is  2  variable  or 
fixed?  Can  you  define  a  constant? 

(6)  When  a  variable  assumes  or  may  assume  every  assignable 
value  between  two  given  constant  values  the  variable  is  said  to 
be  continuous  or  to  vary  in  a  continuous  manner,  in  the  inter- 
val between  the  given  constants. 


64  NUMBERS,  VARIABLES,  FUNCTIONS,  LIMITS 

The  totality  of  values  between  two  given  values  is  called  an 
interval.  If  the  given  values  are  included  the  interval  is  closed, 
if  not  the  interval  is  open. 

If  a  variable  may  not  assume  all  values  in  a  given  interval  it 
is  not  continuous  in  that  interval,  though  it  may  be  continuous 
in'  parts  of  the  interval. 

Suppose  a  bottle  of  ink  stands,  uncorked,  on  the  table  for 
several  days  or  weeks,  and  suppose  it  is  not  disturbed  hi  any 
way.  Is  the  amount  of  ink  in  the  bottle  from  time  to  time  the 
same?  Is  the  quantity  of  ink  constant?  Does  it  vary  con- 
tinuously ? 

Suppose  wheat  is  $1.75  per  bu.  today,  $1.90  yesterday  and 
$2.00  tomorrow.  Is  the  price  of  wheat  constant?  Does  it 
vary  continuously? 

(c)  Any  set  of  numbers  taken  in  some  definite  order  is  called 
a  sequence.     If  there  are  infinitely  many  numbers  in  the  set  the 
sequence  is  called  an  infinite  sequence.     If  there  are  only  a 
finite  number  of  numbers  in  the  set  it  is  a  finite  sequence. 
Thus 

1,    3,    4,    7,    9,     15, 

form  a  finite  sequence.    But 

1,     1.1,     1.11,     1.111,  .  .  . 

form  an  infinite  sequence.  A  sequence  may  increase  or  decrease. 
That  is,  the  successive  numbers  may  be  larger  than  or  smaller 
than  the  preceding,  respectively. 

(d)  Let  x  be  a  variable  and  a  some  constant.     Then  if  a  —  x 
assumes  its  sequence  of  values  in  order,  there  comes  a  stage, 
such  that,  for  all  subsequent  values  of  x,  the  numerical  value  of 
a  —  x  becomes  and  remains  less  than  any  assigned  small  value 
e,  then  a  is  called  the  limit  of  x. 

This  definition  is  expressed  in  symbols  as  a  =  limit  x,  or 
x  — >  a  or  x  =  a.  All  these  forms  may-  be  read  a  is  the  limit  of 
a;  or  a;  approaches  a  as  a  limit. 

A  variable  may  or  may  not  become  equal  to  its  limit,  depend- 
ing on  the  nature  of  the  law  of  its  change. 


A  VARIABLE  65 

The  difference  between  a  variable  and  its  limit  is  a  variable 
whose  limit  is  zero. 

A  variable  whose  limit  is  zero  is  called  an  infinitesimal. 

A  variable  that  may  increase  without  limit  is  said  to  have  no 
limit  or  with  less  propriety  to  have  infinity  for  its  limit. 

(e)  Let  y  be  a  variable  and  x  another  variable.  If  to  every 
value  assigned  to  x  there  corresponds  a  definite  value  assumed 
by  y,  then  y  is  a  function  of  x. 

More  concretely  but  less  exactly  it  may  be  said  that  the 
values  of  y,  the  function,  depend  upon  the  values  of  x,  the  vari- 
able. From  this  notion  we  have  the  terms  dependent  variable 
or  function  and  independent  variable  or  merely  variable.  In 
the  above  definition  x  is  the  independent  variable  and  y  the 
dependent  variable.  In  dealing  with  equations  and  their 
graphs  it  is  customary  to  regard  the  abscissa  as  the  independent 
variable  and  the  ordinate,  the  dependent  variable  or  function. 

38.   When  it  is  desired  to  express  briefly  the  fact  that  y  is  a 
function  of  x  we  may  write: 
y  =  f(%)  (read  y  equals/  of  x  or /function  of  x)  or  y  =  4>(x),  etc. 

This  is  a  notation  used  for  convenience.  Suppose  y  is  defined 
as  a  function  of  x  by  the  expression, 

y=f(x)  =  4x*  -Qx  +  4* 

This  equation  defines  f(x)  for  this  particular  case  and  during 
the  discussion  of  this  function  f(x)  is  understood  as  a  brief  way 
of  writing  the  polynomial  4  x3  —  6  x  +  4.  When  particular 
values  are  substituted  for  x  the  fact  is  indicated  by 

/(3)  =  4  •  33  -  6.3  +  4, 
.    /(-l)=4.(-l)'-6.(-l)  +  4, 
/(a)  =  4  a3  -  6  a  +  4,  etc. 

If  f(x)  is  any  function  of  x  and  if  /(a)  —  /(z)  — *  0,  when 

x  — >  a,  f(x)  is  called  a  continuous  function  of  x  at  the  value 

x  -•  a.     It  is  assumed  that  x  — >  a,  either  by  decreasing  or  by 

increasing  values.     This  is  expressed  by  writing  |  /(a)  —  f(x)  \  — » 0 

*  This  equation  is  called  a  functional  equation  between  x  and  y. 


66  NUMBERS,  VARIABLES,  FUNCTIONS,  LIMITS 

where  the  "|      |"  indicate  the  absolute  value;    that  is,  the 
numerical  value  without  a  plus  or  minus  sign. 

39.  It  is  one  of  the  chief  problems  of  mathematics  to  dis- 
cover and  to  study  functions  of  variables.     This  study  is  most 
easily  carried  on  when  the  function  is  expressed  as  an  equation 
between  the  function  and  the  independent  variable.     Such  an 
equation  is  called  a  functional  equation.     By  studying  the 
equation  the  mathematician  learns  the  character  or  properties 
of  the  function  which  the  equation  expresses. 

When  any  natural  phenomenon  becomes  so  well  known  that 
it  can  be  described  by  a  functional  equation,  it  becomes  a  part 
of  mathematics.  The  mathematician  may  discover  further 
facts  and  peculiarities  of  the  phenomenon.  In  this  way  science 
and  its  applications  have  been  greatly  extended. 

40.  The   difference   between  two   successive   values   of   a 
variable  is  called  an  increment  of  the  variable.     Let  x\,  Xz  be 
two  successive  values  of  the  variable  x.     Then  Xz  —  x^  is  the 
increment  of  x.    The  symbol  Arc  *  is  used  to  represent  the 
increment  of  x.     Thus  Xz  —  Xi  =  Ax.     When  Xz  >  xi}  Ax  >  0. 
When    Xz  <  xi}    Ax  <  0.     Similar    definitions    and    notations 
apply  to  any  variable  and  to  functions. 

41.  Special  forms  and  limits.  —  Theorems. 

a 


I.   Lim 


II.   Lim 


III.   Lim 


IV.  Lim 


=  0,  where  a  is  a  definite  number. 


=  oo ,  where  a  is  a  definite  number. 


=  0,  where  a  is  a  definite  number  not  zero. 


=  oo ,  where  a  is  a  definite  number*  not  zero. 


X 

The  student  can  easily  satisfy  himself  concerning  the  rea- 
sonableness of  these  theorems  by  arithmetical  methods.  For 
example,  consider  the  sequence  of  fractions  with  the  same 
numerators  but  with  increasing  denominators, 


*  Read  delta  x. 


SPECIAL  FORMS  AND  LIMITS 


67 


From  our  knowledge  of  division  in  arithmetic  it  is  evident  that 
the  successive  fractions  are  smaller  and  smaller  in  value  as  we 
proceed  with  the  sequence.    That  is,  they  are  nearer  and  nearer 
zero.    They  are  approaching  zero. 
Below  are  given  the  usual  proofs  of  the  above  theorems. 

42.   I.   The  limit  of   -  as  x  becomes  infinite  is  zero,  that  is, 


lim 
x 

Suppose 


=  0,  where  a  is  a  definite  number. 


<  e    or    \-\  <  e 
\x\ 


where  e  is  an  arbitrarily  small  number,  not  0.    Then 

\a\<e\x\     and     |*|>^« 
Therefore,  for  any  assigned  value  of  e,  e  ?*  0,  we  can  calculate 


a  value  of  x  such  that 


increasing  values  of  x, 


<  €.    It  follows  that  for  indefinitely 


satisfies  the  definition  of  limit  of  a 


variable  and  has  zero  for  its  limit. 


II.    Lim 


=  oo,  where  a  is  a  definite  number. 


It  will  be  sufficient  to  show  that  if  a;  is  chosen  sufficiently 


large  the  inequality 


>  M  can  be  satisfied,  however  large 


M  is  chosen.     For  multiplying  both  sides  by  |  a|  gives 

\x\  >\a\M. 
It  is  only  necessary  then  to  choose  x  >  \  a  \  M  in  order  to  ensure 


the  inequality 


III.    Lim 


>  M ,  however  large  M  may  be. 


=  0,  where  a  is  a  definite  number  not  zero. 


It  will  be  sufficient  to  show  that  if  a;  is  chosen  sufficiently 


68  NUMBERS,  VARIABLES,  FUNCTIONS,  LIMITS 

\x\ 

small  the  inequality,  r  -4  <  e,  however  small  e  be  chosen,  will  be 
I  al 

satisfied.     Multiplying  both  sides  by  |  a  |  gives 

I  a;  |  <  e  |a|. 

Now  if  a  and  e  are  given  x  is  determined  at  once  so  as  to  satisfy 

\x\ 
the  condition  ]—[<«. 

a 


IV.  Lim 


=  oo,  where  a  is  a  definite  number  not  zero. 


x->0 

It  will  be  sufficient  to  show  that  if  x  be  chosen  sufficiently  smaD 

the  inequality  t-  4  >  M ,  however  large  M  is  chosen,  holds. 
\x\ 

Multiplying  both  sides  by  |  x  \  gives 

\a\>M\x\. 
Dividing  now  by  M, 


This  value  of  x  will  ensure  the  first  inequality. 

43.  It  is  often  desirable  to  know  the  limit  approached  by 
an  expression  when  the  variable  approaches  a  given  value.  The 
preceding  theorems  are  useful  for  this  purpose. 

x  —  2 

1.   What  is  the  limiting  value  of  -5 — : — -..  when  x— >2. 

x2  —  4x  +  4' 

By  direct  substitution  of  x  =  2  in  the  function  the  result  is  0/0. 
This  result  can  have  no  meaning.  But  it  is  noticed  that  this 
expression  is  not  in  its  lowest  terms.  For 

x-2  1 


s2-4z  +  4      x-2 

Now  as  x  =  2,  the  result  is  oo ,  by  IV.    The  expression  0/0 
may  be  assigned  any  value.     For,  write 

x      j 
-  =  k 

y 


NUMBER  PAIRS  69 

and  let  x  — »  0  and  y  — >  0  in  such  a  way  that  this  equation  always 
holds.  Multiplying  by  y  gives 

x  =  ky. 

This  equation  is  satisfied  even  when  x  =  0  and  y  =  0.  But 
k  was  arbitrary,  therefore  §  may  have  any  value  and  is  indeter- 
minate. 

2.   What  is  the  limit  of  — ^ 5— ^ —  as  x— »oo.     By  direct 


substitution  the  result  is  ^/^  which  is  equally  as  indeter- 
minate 0/0.  But  if  numerator  and  denominator  be  divided  by 
y?  the  expression  becomes, 


x      x 


Now  as  x  —  >  oo  ah1  the  terms  of  the  numerator  become  0,  by  I. 
The  denominator  becomes  1.  Therefore  the  value  of  the 
fraction  becomes  0,  by  III. 

x  —  1 

3.  What  is  the  limit  of  J       -^  ,  when  x  —  >  1  ? 

o^  —  l' 

4.  What  is  the  limit  of  z2  —  6  x  +  2  when  x  —  »  0,   x  —  »  2, 

x  —  >  10,  respectively  ? 

^ 

5.  What  is  the  limit  of  ex,  when  x  —  >  oo  ? 

6.  What  is  the  limit  of  x  +  -,  when  x  —>  oo  ?    Whenz—  >0? 

x 

x  —  2 

7.  What  is  the  limit  of  -^  --  r,  when  z  —  >  0? 

X    ^~  X 

8.  What  is  the  limit  of  e~x,  when  x  —  >  0  ?    When  x  —  >  oo  ? 
44.   It  should  be  noted  that  in  much  of  the  work  of  previous 

chapters  we  were  concerned  with  number  pairs.  We  selected 
one  number  of  a  pair  and  found  the  other  by  observation  or 
calculation.  This  correspondence  of  number  pairs  satisfies  the 
definition  of  function.  Suppose  (xi,  t/i),  (a*,  3/2),  (xa,  ya),  .  .  .  , 
(xn,  yn)  be  a  set  of  number  pairs,  such  that  as  a  variable  x 


70  NUMBERS,  VARIABLES,  FUNCTIONS,  LIMITS 

* 

assumes  the  values  Xi,  x^,  .  .  .  ,  xn,  another  variable  y  assumes 
the  corresponding  values,  y\,  y%,  .  .  .  ,  yn,  the  variable  y  is  to 
be  regarded  as  a  function  of  the  variable,  x,  over  the  given  set 
of  values. 

If  each  pair  of  values  of  x  and  y  be  regarded  as  the  coordinates 
of  a  point,  a  curve  drawn  through  these  points  represents  more 
or  less  approximately  the  function  in  question.  In  fact  the 
curve  may  be  regarded  as  defining,  approximately,  a  function, 
a  few  of  whose  values  are  known,  viz.,  the  points  used  in  draw- 
ing the  curve.  By  measuring  the  abscissas  and  ordinates  of 
other  points  of  the  curve  we  may  determine  any  number  of 
other  values  of  the  variable  and  the  corresponding  values  of 
the  function,  approximately.  It  is  seen  that  the  graphic 
representations  of  previous  chapters  represent  or  define,  ap- 
proximately at  least,  functions,  whether  we  know  the  functional 
equations  or  not.  In  some  cases  we  were  able  to  discover  an 
equation  from  the  graph. 

It  is  essential  to  progress  in  mathematics  and  its  applications 
that  we  recognize  the  use  of  the  graph  as  a  means  of  studying 
functions  and  of  discovering  the  corresponding  functional 
equations,  and  that  we  learn  to  study  any  function  by  means 
of  both  its  equation  and  its  graph. 

In  the  next  chapter  we  shall  study  a  class  of  functions  which 
have  a  wide  range  of  application,  not  only  in  mathematics  but 
in  science  and  engineering  as  well.  In  later  chapters  we  shall 
study  still  other  kinds  of  functions. 


CHAPTER  VIII 
THE  TRIGONOMETRIC  FUNCTIONS 

45.  Problem.  —  It  is  desired  to  know  the  height  of  a  tree, 
BC.  It  is  found  that  the  distance  AC  and  the  angle  CAB  can 
be  measured.  From  this  data  it  is  desired  to  find  the  height  of 
the  tree,  but  these  measurements  alone  will  not  be  sufficient 


FIG.  20. 

for  the  purpose.  The  additional  measurements  AC'  B'C' 
are,  therefore,  taken.  This  is  done  by  means  of  the  upright 
pole  B'C'  such  that  AB'B,  the  line  of  sight,  is  a  straight  line. 
Now  from  the  similar  triangles  AC'B'  and  A CB  the  proportion 
B'C'  BC 


or 

(2) 


AC'       AC1 


B'C' 

BC  =  ~T7if  '  AC 


can  be  written.  Equation  (2)  shows  that  BC  depends  on  A C 
and  the  ratio  B'C' /AC'.  This  ratio  evidently  depends  in  some 
way  on  the  angle  CAB  =  6.  The  frequent  occurrence,  in 
science  and  engineering,  of  situations  similar  to  this  caused 

71 


72 


THE  TRIGONOMETRIC  FUNCTIONS 


mathematicians  to  search  for  the  functional  relation  of  the  ratio 
to  the  corresponding  angle.  By  careful  measurements  and 
calculations  the  values  of  the  ratio  and  the  corresponding  angle 
have  been  tabulated  for  convenience  in  solving  problems. 

The  ratio  B'C'/AC'  is  called  the  tangent  ratio  of  the  angle  0. 
Having  a  table  of  such  ratios  for  different  angles  it  is  easy  to 
calculate  the  height  of  the  tree  from  the  original  measurements 
of  AC  and  the  angle  6  =  CAB.  Equation  (2)  may  now  be 
written 

(3)  BC  =  AC  (tangent  of  6) 
or  more  briefly 

(4)  BC 


FIG.  21. 

In  this  chapter  we  shall  study  the  theory  and  use  of  the 
tangent  ratio  and  other  related  ratios.  The  methods  developed 
will  furnish  ways  of  solving  problems  of  the  highest  importance 
in  science  and  engineering. 

46.  An  angle  will  now  be  regarded  as  generated  by  the 
rotation  of  a  straight  line  about  one  of  its  points  from  some 


HESSLER'S  DEFINITIONS 


73 


initial  position  to  some  final  or  terminal  position.  The  angle 
will  be  included  between  lines  lying  in  these  two  positions  with 
its  vertex  at  the  point  of  rotation.  For  the  purpose  of  formulat- 
ing the  definitions  and  fundamental  relations  the  line  OX  in 
Fig.  21  will  be  taken  as  the  standard  initial  line  or  position. 
Let  OP  rotate  from  OX,  about  0,  counterclockwise.  Let  OP', 
OP",  OP"7,  OPIV  be  successive  positions  chosen  as  OP  rotates. 
These  lines  are  the  terminal  lines  of  the  angles,  a\,  a2,  a3,  at, 
respectively,  as  shown  in  the  figure. 

Counterclockwise  rotation  is  regarded  as  positive  and  clock- 
wise as  negative.  This  convention  gives  rise  to  positive  and 
negative  angles  to  correspond.  The  angle  a&  is  negative. 

47.  Hessler's  definitions  of  the  trigonometric  functions 
(ratios)  are  as  follows:  (See  Fig.  22.) 


Name. 

Symbol. 

a, 
Quad.  I. 

Quad2.  II. 

Quad!  III. 

Quad!  IV. 

sine  of  a  

sin  a 
cos  a 
tana 
cot  a 

sec  a 

CSC  a 

vers  a 
covers  c 

<+>! 

(+)- 

=  1  —  CO 

*  =  1  —  sir 

(     ^ 
(     }h2 

X2 
*^2 

<->  1 

S« 
1  a 

.hi 

(+)  — 

(-)f-; 

9t 

(-)- 

cosiile  of  a  

tangent  of  a  

cotangent  of  a  

secant  of  a  

cosecant  of  a  

versed  sine  of  a  
coversed  sine  of  a  ... 

Note  the  signs  of  the  ratios  in  the  different  quadrants.    These 
depend  on  the  signs  of  x  and  y  in  the  various  quadrants. 

The  above  table  and  figure  must  be  memorized  as  a  basis  of 
future  work. 

The  haversine  is  defined  as  follows: 

hav  x  =  |  vers  x  =  $  (1  —  cos  x). 

By  use  of  tables  of  natural  haversines  and  their  logarithms  the 
solution  of  many  of  the  problems  in  nautical  astronomy  is 
greatly  simplified. 


74 


THE  TRIGONOMETRIC  FUNCTIONS 


An  angle  is  said  to  be  in  or  to  lie  in  that  quadrant  in  which  its 
terminal  line  lies.    Thus  «i  is  in  the  first  quadrant,  and  «3  is  in 


Quad.  II 


the  third  quadrant.  Functions  of  negative  angles  are  defined 
exactly  as  for  positive  angles.  Thus  sin  a5  =  yt/hi,  etc. 

1.  In  what  quadrant  is  the  angle 
140°?    210?    85°?    240? 

2.  What  is  the  sign  of  each  function 
of  each  angle  in  Ex.  (1)  above  ? 

3.  Draw  an  arc  of  90°  with  a  radius 
10  cm.    Divide  the  arc  into  10°  arcs. 
Measure  the  coordinates  of  the  points 
of  division.    Divide  each  ordinate  by 
the  radius,  10  cm.,  and  tabulate  the 
quotients.    Divide   each   abscissa  by 
the  radius  and  tabulate.     Divide  each 

ordinate  by  the  corresponding  abscissa.  Arrange  all  these  ratios 
in  a  table  with  the  corresponding  angles  and  compare  them 
with  the  values  given  in  the  table  of  sines,  cosines  and  tangents 
for  the  same  angles,  respectively,  in  the  back  of  the  book. 


FIG.  23. 


FUNDAMENTAL  FORMULAS  75 

48.  From  the  definitions  of  47  and  the  Pythagorean  theorem 
the  following  fundamental  formulas  which  hold  in  all  the  quad- 
rants are  derived: 


(1) 

(x\2 
s)  + 

fy\2     x2 
(h)  =  h*  ' 

+-r2  =  §  =  L       •'•    cos2  a  +  sin2  a  =  1.* 

nr       hr 

(2) 

y 

X 

1 

x 

cot  a 

y 

(3) 

h 

i 

sec  a  =  or  cos  a  sec  a  =  1. 

x 

X 

cos  a 

h 

h 

1 

1 

fA\ 

w 

y 

y 

sin  a 

h 

ft(\ 

(h 

V    i-' 

i2  -  x2  _  y2                _2  _                _2  ' 

(6) 


h      y  sin  a 

(7)  -  =  -  .    .'.    -  =  tan  a. 

x      x  cosa 

h 

The  formulas  in  the  right-hand  column  must  be  memorized. 

1.  Given  sin  A  =  1/2,  find  cos  A  and  tan  A.     Construct  a 
right  triangle  whose  hypotenuse  is  2  and  whose  vertical  leg  is  1, 
as  in  Fig.  24.    Calculate  x  =  A/22  -  I2  =  V§  =  1.732. 

1  7*32 
cos  A  =  -     -  =  0.866,  by  the  definition,  47. 

£i 

sin  A        0.5        _  ___ 
tan  A  =  -     T  =  jr^^  =  0.577. 
cos  J.      0.866 

2.  Given  sec  A  =  3,   find  cos  A.   and  sin  A.    Construct  a 

*  The  symbol  sin2  a  is  used  for  (sin  a)2,  being  more  convenient. 
notations  are  used  for  the  other  ratios. 


76 


THE  TRIGONOMETRIC  FUNCTIONS 


right  triangle  whose  hypothenuse  is  3  and  horizontal  leg  is  1  as 
in  Fig.  25.      Calculate  y  =  \/32  -  I2  =  Vs  =  2  \/2  =  2.828. 

y      2.828 


„„„    A          •"  —  *-  —  O  ^33          sin  4    —  y   — 
COS  ^x   ==  ~~"         ~~  "~~  U«OOO*         bill  £\.   —  y  — 

/I         O  /v  O 

3.  Given  sin  A  =  f ,  calculate  cos  A,  tan  A, 
and  sec  A. 

4.  Given  tan  A  =  *,  calculate  sin  A  and 
cos  A. 

5.  If  one  leg  of  a  right  triangle  is  24  and 
the  hypotenuse  30,  find  all  the  functions  of 
the  angle  adjacent  the  given  leg. 

6.  One  leg  of  a 
right  triangle  is  half 
the  hypotenuse. 
Find  all  the  func- 
tions of  the  angle 


,*«,«_ 
=  0.9427. 


opposite  the  leg. 


FIG.  24. 


£C=1 

FIG.  25. 


7.  If  sec  0  =  1.5,  find  tan  6  and  sin  6. 

8.  If  tan  6  =  2.5,  find  sin  6  and  cos  6. 

9.  The  hypotenuse  of  a  right  triangle  is  12,  the  base  is  8,  find 
the  sine  and  tangent  of  the  angle  opposite  the  base. 

10.  If  the  tangent  of  A  is  1,  find  the  secant  of  A  and  the 
sine  of  A. 

In  order  that  the  student  shall  become  sufficiently  familiar 
with  the  seven  fundamental  formulas  and  their  use  the  exercises 
below  should  be  solved.  The  given  equations  are  identities. 
Either  or  both  sides  are  to  be  modified  by  various  substitutions 
from  the  fundamental  formulas  so  that  both  sides  shall  appear 
to  be  identically  equal. 

1.   Cos  A  tan  A  =  sin  A. 

Write  this  in  the  form,  using  Eq.  7, 

i    sin  A 

cos  A  •  —    -r  =  sin  A. 
cos  A 


IDENTITIES  77 

Reducing  the  left  side  this  becomes 
sinA  =  sin  A 

which  is  known  to  be  identically  true  for  all  values  of  A. 

2.  Sec  A  sin  A  =  tan  A.     (Use  equations  3,  7.) 

3.  Cos  A  esc  A  =  cot  A. 

4.  (sin  A  —  cos  AY  =  1  —  2  sin  A  cos  A.     (Expand  and  use 
Eq.  1.) 

5.  sin  A  cot  A  =  cos  A. 

6.  esc  A  tan  A  =  sec  A. 

7.  cos  A /(sin  A  cot2  A)  =  tan  A . 

8.  (sin3  A  —  cos3  A)  =  (sin  A  —  cos  A)  (1  +  sin  A  cos  A). 

9.  (acosA  +6sinA)2+  (asinA  -6  cos  A)2  =  a2  +  b*. 

10.  sec2  A  +  esc2  A  =  tan2  A  +  cot2  A  +  2. 

11.  cos  A/(l  —  tan  A)  +  sin  A/(l  —  cot  A)  =  sin  A  +  cos  A. 

12.  cot8  A  -  cos2  A  =  cos2  A  cot2  A . 

13.  tan2  A  -  sin2  A  =  sin4  A  sec2  A. 

14.  sec  A  +tanA  =  cosA/(l  —  sin  A). 

15.  (1  +  sin  A  +  cos  A)2  =  2  (1  +  sin  A)  (1  +  cos  A). 

16.  cot4  A  +  cot2 A  =  esc4  A  —  esc2  A. 

17.  (sin  A  +  esc  A)2  +  (cosA  +  sec  A)2  =  tan2A  +  cot2  A  +7. 

18.  sin  A  (tan  A  —  1)  —  cos  A  (cot  A  —  1)  =  sec  A  —  esc  A. 

19.  tan2  A  -  cot2  A  =  sec2  A  esc2  A  (sin2  A  -  cos2  A). 

20.  2  vers A  +  cos2  A  =  1  +  vers2  A. 

21.  cos2  A  (1  + tan2  A)  =  1. 

22.  (sec2  A  -  1)  (esc2  A  -  1)  =  1. 

23.  tan  A  +  cot  A  =  sec  A  esc  A . 

24.  einA/cscA  +  cos  A/sec  A  =  1. 

25.  sec2  A  -sec5  A  sin2  A  =  1. 

26.  (tan  A  -  l)/(tan  A  +  !)  =  (!-  cot  A)/(l  +  cot  A). 

27.  (tan  A  +  cot  A)2  =  sec2  A  +  esc2  A . 

28.  (secA-cscA)/(secA+cscA)  =  (tanA-l)/(tanA+l). 

29.  (esc  A  -  cot  A)2  =  (1  -  cos  A)/(l  +  cos  A). 

/*«2         rt/2 

30.  Show  that  —2  +  ^  =  1,  if  x  =  a  cos  A,  j/  =  6  sin  A. 

a       o 


78 


THE  TRIGONOMETRIC  FUNCTIONS 


49.   Functions  of  angles  at  the  quadrant  limits : 

1.  Sin  0°  and  cos  0°. 

In  Fig.  26  as  OP  moves  toward  OX,  the  angle,  0-»0°. 
At  the  same  time  the  ordinate,  y— »0  and  x— >h.  We  fire, 
therefore,  justified  in  concluding 

11       0 
Lim  sin  0  =  Lim  f  =  r  =  0. 

9— >0°  y— »0'l        n 

:.     sin  0°  =  0. 

By  similar  reasoning, 

x      h 

Lim  cos  0  =  Lim-r  =  ,-  =  1. 
e— »o*  x—*hh      ft 

:.     cos  0°  =  1. 


-   i  "•  — i — j-  f^^  •        '      '       '    '        'I          A 

y\ ——•        Ojfcfc'~~  -jV  \—0 

•— pt~~~  I   \  ~"° * 

/     \ 


.       Y'  \ 

FIG.  26. 

In  Fig.  26  OP  =  h,  OP'  =  h',  etc. 

2.   Sin  90°  and  cos  90° :  By  reasoning  analogous  to  the  above 
Lim  sin  0  =   Lim  \j  =  77  =  1. 
sin  90°  =  1. 


QUADRANTAL  LIMITS  79 

Again, 

x'      0 
Lim  cos  0  =  Lim  rj  =  -n  =  0. 

g  _>  90°  z'  — » 0  h  h 

:.     cos  90°  =  0. 
3.   Sin  180°  and  cos  180°: 

v"       0 
Lim  sin0  =  Lim  rr/  =  777  =  0. 

0— »180°  y"->otl  n 

.-.     sin  180°  =  0. 

x"       —h" 
Lim    cos  6  =    Lim    ^7,  =  -777-  =  —  1. 

0-+180"  x"-^-^'^  ft 

/.     cos  180°  =  -1. 

Note.  —  It  is  evident  that  each  of  the  limits  above  is  the  same 
whether  the  respective  variables  increase  or  decrease  toward 
their  limits. 

1.  Derive  by  a  method  similar  to  the  above  sin  270°  =  —  1, 
cos  270°  =  0;  sin 360°  =  0,  cos  360°  =  1. 

2.  From  the  values  of  the  sines  and  cosines,  by  use  of  the 
fundamental  formulas,  obtain  the  values  of  the  tangent,  co- 
tangent, secant  and  cosecant  of  0°,  90°,  180°,  270°,  360°. 

50.  As  was  implied  in  46,  negative  angles  and  their  functions 
must  be  considered.  It  is  easily  seen  that  for  every  positive 
angle  there  exists  a  negative  angle  of  equal  magnitude.  The 
coordinates  of  P'  (Fig.  27)  determine  the  functions  of  —6  in 
the  same  way  that  the  coordinates  of  P  determine  the  func- 
tions of  0. 


Now 


-y 

h 


Hence  |  sin  0 1  =  |  sin  ( — 0)  |  and 

(8)  sin  (-8)  =  -sin  8. 

Since  x  and  h  are  the  same  for  —9  as  for  0,  it  follows  that 

(9)  cos  (-8)  =  cos  8. 

1.  By  use  of  the  formulas  of  47  derive  tan  (—0),  cot  (—0), 
sec  (—0)  and  esc  (—0)  in  terms  of  the  same  named  functions 
of  0. 


80 


THE  TRIGONOMETRIC  FUNCTIONS 


2.  How  can  sin  (—35°),  cos  (—20°)  be  determined  from  a 
table  of  sines  and  cosines  where  only  positive  angles  are  con- 
sidered ? 

Y 


FIG.  27. 

51.  If  for  any  position  of  OP,  Fig.  28,  where  P  is  the  point  P 
(x,  y)  a  line  OQ  be  drawn,  where  Q  is  the  point  Q  (x\,  yi),  where 
x  =  yi>  y  =  xi>  then  the  angle  XOQ  will  be  the  complement  of 
the  angle  XOP.  From  the  definitions  of  47  and  the  above 
construction  it  follows  that  sin  XOP  =  cos  XOQ  or 


sin  a  =  cos  (90°  —  a), 
tan  a  =  cot  (90°  -  a), 


(10) 
and 
(11) 
and 
(12)  sec  a  =  esc  (90°  -  a). 

These  equations  and  the  method  of  demonstration  apply  to  all 
quadrants.  These  formulas  will  be  named  the  complement 
relations. 


REDUCTION  FORMULAS 


81 


1.  From  sin  30°  =  0.5,  find  cos  60°;  From  sec  45°  =  1.414, 
find  esc  45°. 

2.  From  sin  60°  =  0.866,  find  cos  30°;  From  tan  45°  =  1, 
find  cot  45°. 


FIG.  28. 

3.  From  cos  120°  =  -0.5,  find  sin  (-30°). 

4.  From  tan  60°  =  V3,  find  cot  30°. 

5.  The  legs  of  a  right  triangle  are  x  =  5,  y  —  7.    Calculate 
all  the  functions  of  the  angle  opposite  the  longest  leg. 

6.  By  use  of  formulas  of  51  obtain  all  functions  of  the  angle 
opposite  the  shortest  leg  from  the  values  calculated  in  Ex.  5. 

52.  To  reduce  the  functions  of  any  angle  to  the  same  named 
functions  of  an  angle  less  than  90°. 

1.   Consider     90°  <  a  <  180°.      Construct     the    triangles 
POM  and  P'OM',  Fig.  29,  so  that 

h  =  h',    a'  =  A,     -x  =  xf,    then  y  =  y',    A  =  (180°  -  a)  =  a'. 
Then  sin  a  =  y'/h'  =  y/h  =  sin  a'. 

(13)  /.     sin  a  =  sin  (180°  -  a). 


82  THE  TRIGONOMETRIC  FUNCTIONS 

Also,  since  —  x  =  x', 

(14)  cos  a  =    ,  =  -      =  -cos  (180°  -  a). 


FIG.  29. 

2.  Consider  180°  <  a  <  270°.     Construct  the  triangles  POM 
and  P'OM'  (Fig.  30)  so  that  A  =  ar,  y  =  -y',  x  =  -x',  then 

and    a'  =  a  -  180°.     Evidently, 

sin  a  =  y/h  =  —y'/hr  =  —  sin  a/. 

(15)  .'.     sin  a  =  -sin  (a  -  180°). 

Also,        cos  a  =  x/h  =  —x'/hr  =  —  cos  a'. 

(16)  /.  cos  a  =  -  cos  (a  -  180°). 

3.  Consider    270°  <  a  <  360°.      Construct    the    triangles 
POM  and  P'OM  (Fig.  31)  so  that 

A  =  a',    y  =  —yf,    x  =  x',     then  h  =  h', 
and       a'  =  360  -  a  =  A.     Evidently, 

sin  a  =  y/h  =  —y'/h'  =  —sin  a'. 

(17)  /.     sin  a  =  -sin  (360°  -  a), 
and  cos  a  =  x/h  =  x/h'  =  cos  a'. 

(18)  .'.     cos  a  =  cos  (360°  -a ). 


REDUCTION  FORMULAS 


83 


FIG.  30. 


FIG.  31. 


84 


THE  TRIGONOMETRIC  FUNCTIONS 


4.   Consider  90°  <  a  <  180°  and  a  —  90°.     Construct   the 
triangles  POM  and  P'OM'  so  that 

y  =  xf,    x  =  -y',     then  a'  =  a  -  90°. 


Then 

(20) 

Again 


FIG.  32. 


sin  a  =  y/h  =  x'/h'  =  cos  a'. 
sin  a  =  cos  (a  —  90°). 
cos  a  =  x/h  =  —  y'/h'  —  —  sin  a'. 
cos  a  =  —sin  (a  —  90°). 


FIG.  33. 


As  an  exercise  let  the  student  derive  from  these  results  the 
tangent  of  a,  under  all  the  above  cases. 


ADDITION  THEOREMS 


85 


5.  Consider  360  <  a.  It  is  evident  that  if  360°  be  added  to 
any  angle  a,  the  terminal  line  will  coincide  with  that  of  a.  It 
follows  that  the  values  of  the  defining  ratios  of  the  functions  of 
a  +  360°  will  be  identical  with  those  of  a.  It  is  evident  then 
that 

(22)  sin  a  =  sin  (360  +  a), 

(23)  cos  a  =  cos  (360  +  a) , 

and  similarly  for  all  the  functions.     It  is  equally  evident  that 
360°  may  be  subtracted  from  a  without  affecting  the  values  of 
the  functions  of  a. 
63.   Addition  theorems.  —  The  formulas 

(1)  sin  (A  +  B)  =  sin  A  cos  B  +  cos  A  sin  B, 

(2)  cos  (A  -f  B)  =  cos  A  cos  B  —  sin  A  sin  B 

are  known  as  the  addition  theorems  for  the  sine  and  cosine 
respectively. 

Y 


FIG.  34. 

To  prove  (1),  consider  either  position  of  Q  (Fig.  34a  and  346), 
where  0°  <  A  <  90°  and  0°  <  B  <  90°.    Then  A  +  B  <  180°. 
sin  (A  +  fi)  =  NQ/OQ  =  MP/OQ  +  LQ/OQ. 


86 


THE  TRIGONOMETRIC  FUNCTIONS 


But  MP  =  OP  sin  A  and  LQ  =  PQ  cos  A,  since  angle  LQP 
A.    Substituting  these  values  in  the  above  equation  gives 


But 


sin  (A  +  B]  =  -Qfi  sin  A  +  ~  cos  A. 

OP  ,    PQ 

•      =  cos  B    and  =  am  B. 


Substituting  these  values  in  the  last  equation  gives 
(24)  sin  (A-\-B)  =  sin  A  cos  B  +  cos  A  sin  B. 

This  is  equation  (1). 


Fia.  34. 


To  prove  (2)  consider  the  same  figure: 

i  A       m      ON     OM     NM 
cos  (A  +  B)  = 


OM 


LP 
OQ 


OQ       OQ       OQ       OQ 
But  OM  =  OP  cos  A    and    LP  =  QP  sin  A. 

Substituting  these  values  in  the  above  equation  gives: 

QP 


But 


cos  (A  +  B)  =  ^  cos  A  —  ^~  sin  A. 

OP  PQ 

TTK  =  cos  B     and    T^T\  =  sin  B. 


ADDITION  THEOREMS 


87 


Substituting  these  values  in  the  last  equation  gives 
(25)  cos  (A  +  B)  =  cos  A  cos  B  —  sin  A  sin  B. 

This  is  equation  (2). 

These  proofs  may  be  extended  to  angles  of  any  magnitude  by 
the  use  of  52.  For  a  >  90°,  sin  a  can  be  expressed  in  terms  of 
sin  a'  where  a'  <  90°.  Therefore,  when  A  >  90°  and  B  >  90° 
and  A  +  B  >  180°  the  sines  and  cosines  of  A  and  B  can  be 
expressed  in  terms  of  like-named  functions  of  angles  less  than 


Fia.  35. 

90°,  say  A'  and  B',  respectively.  Corresponding  to  A  +  B  > 
180°  there  will  be  A'  +  B'  <  180°.  Then  sin  (A  +  B)  can  be 
expressed  in  terms  of  sin  (A'  +  B').  The  generality  of  the 
equations  (1),  (2)  for  all  angles  is  inferred.  The  proof  may  also 
be  generalized.  If 

90°  <  A  <  180°,    90°  <  B  <  180°,     180°  <A  +  B  <  360°. 
Then  in  the  figure 

sin  (A  +  B)  =  -sin  (A'  +  B'}  =  y'/h'  =  -y/h, 
A  +  B  -  180°  =  A'  +  B'  =  (A  -  90°)  +  (B  -  90°), 
where    0°  <  A'  <  90°,    0°  <  B'  <  90°,    0°  <  A'  +  B'  <  180°. 

/.     sin  (A  +  B}  =  -sin  (A'  +  B'), 
This  case  has  been  proved,  since  A'  +  B'  <  180. 


88  THE  TRIGONOMETRIC  FUNCTIONS 

It  is  known  that: 

sin  (A  -  90°)  =  -cosA  =  sin  A', 
cos  (A  —  90°)  =  sin  A  =  cos  A', 
sin  (B  -  90°)  =  -cos 5  =  sin Br, 
cos  (B  -  90°)  =  sin  B  =  cos  B'. 
From  these 
tan(A+B)=  -sin  (A' 

.'.     sin  (A  +  B)  =  sin  A  cos  B  +  cos  A  sin  5. 

This  establishes  the  theorem  for  all  values  of  A  and  B  so  long  as 
A  +  B  <  360°.  As  an  exercise  let  the  student  establish  the 
equation  (2)  for  the  same  values  of  A,  B. 

In  a  similar  manner  the  proof  may  be  extended  to  the  case 
of  A  -f  B  >  360°.  The  addition  theorems  are  general. 

The  addition  theorems  for  the  other  functions  can  be  easily 
deduced  from  those  of  the  sine  and  cosine  by  algebraic  methods. 
Thus  for  tan  (A  +  B),  write 

,  A    .    m      sin  (A  +  B)      sin  A  cos  B  +  cos  A  sin  B 

tan  (A  +  B)  =  — .  .    ,    p;  =  -  — = r—       — 5- 

cos  (A  -f-  J5)      cos  A  cos  B  —  sin  A  sin  .B 

Dividing  numerator  and  denominator  of  the  last  fraction  by 
cos  A  cos  B  gives 

sin  A  cos  B      cos  A  smB 

/  A  i  n\   cos  A  cos  B   cos  A  cos  5 
tan  (A  +  5)  = 


cos  A  cos  B      sin  A  sin  B 
(26)      /.     tan  (A  +  B)  = 


cos  A  cos  J?   cos  A  cos  5 
tan  yi  +  tan  B 


1  —  tan  A  tan  B 

for  all  values  of  A  and  B. 

1.  By  use  of  the  definitions  of  functions  of  negative  angles 
derive  from  (24),  (25),  (26), 

(27)  sin  (A  —  B)  =  sin  A  cos  B  —  cos  A  sin  B. 

(28)  cos  (A  —  B}  =  cos  A  cos  B  +  sin  ^4  sin  B. 

/nr\\  /  A  tan  A  —  tan  B 

(29)  tan  (A  —  B)  =  r-r-r 77 &• 

1  +  tan  ^4  tan  B 


THE  SOLUTION  OF  TRIGONOMETRIC  EQUATIONS      89 

2.  If    sin  A  =  0.5    and    sin  B  =  0.25,    find    sin  (A  +  B). 
cos  (A  +  B),  tan  (A  +  B),  sin  (A  -  £),  cos  (A  -  B),  tan  (A  -  B). 

Note.  —  Cos  A  and  cos  B  must  first  be  found.  Then  sub- 
stitute in  the  addition  theorems. 

3.  By  use  of  the  addition  theorems  derive: 

(30)  sin  (90°  +  A)  =  cos  A. 

(31)  cos  (90°  +  A)  =  -  sin  A. 

(32)  sin  (90°-^)  =  cos^. 

(33)  cos  (90°-^)  =  sin,4. 

(34)  sin  (180°  -A)  =  sin  A. 

(35)  cos  (180°  -  A)  =  -cos  A. 

(36)  sin  (A  -  90°)  =  -cos .4. 

4.  By  use  of  the  addition  theorem  for  A=BorA+A  = 
2  A,  derive 

(37)  sin  2  A  =  2  sin  A  cos  A, 

(38)  cos  2  A  =  cos2  A  —  sin2  A 

=  2  cos2 ,4  -  1 
=  1-2  sin2 .4. 

5.  By  use  of  the  addition  theorems  with  A  +  2  A  =  3  A, 
derive  expressions  for  sin  3  A  and  cos  3  A  in  terms  of  sin  A  and 
cos  A. 

6.  Given  sin  A  =  0.6,  find  sin  2  A,  cos  2  A,  sin  3  A,  cos  3  A. 

7.  Given  sin  A  =  I ,  sin  B  =  f ,  find  sin  (A  +  B)  cos  (A  +  B) 
tan  (A  +  5). 

8.  Given  cos  2  A  =  0.866,  find  cos  A  and  sin  A. 

M>fe.  —  Write  cos  2  4  =  2  cos2  A  -  1  =  0.866,  and  solve 
the  equation  for  cos  A. 

9.  Given  tan  2  A  =  1.5,  find  tan  A  and  cos  A. 

10.  Given  tan  A  =  0.8,  find  tan  (180  +  A),  tan  (180  -  A). 

11.  From  sin  45°  =  0.7071,  find  sin  22°  30'. 

54.  In  the  solution  of  problems  an  unknown  angle  often 
occurs  through  one  of  its  functions,  say  a  sine  or  cosine.  It  is 
then  necessary  to  solve  the  equation  to  obtain  the  angle.  To  do 
this  the  function  of  the  unknown  angle  is  regarded  as  the 
unknown  quantity  in  the  equation  instead  of  the  angle  itself. 


90  THE  TRIGONOMETRIC  FUNCTIONS 

When  the  function  has  been  found  the  corresponding  angle  can 
be  found  from  the  table.  To  illustrate,  suppose  there  is  given 
1  +  sin  A  =  f ,  from  which  to  find  A.  Transposing  and 
simplif ying  give 

sin  A  =  \. 

From  the  table,  A  is  found  to  be  30°.  By  use  of  formulas  of  52 
sin  30°  =  sin  (180°  -  30°)  =  sin  150°.  Hence  150°  is  another 
value  of  A  which  satisfies  the  equation. 

Instead  of  finding  A  in  degrees  it  may  be  desirable  to  use  A 
merely  as  an  angle  belonging  to  its  sine,  |.  For  this  purpose 
several  notations  are  used.  Thus 

A  =  angle  whose  sine  is  £,     A  =  arc  sin  \,    A  =  sin"1  ^, 
(read  inverse  sine  £)  all  mean  the  same  thing.     Either  of  the 
equations 

A  =  arc  sin  x 
and 

x  =  sin  A 

implies  the  existence  of  the  other.  Similar  notations  apply  to 
all  the  functions  of  the  angle. 

1.  What  is  the  value  of  arc  sin^?    Of  arc  sin  1?    Of  arc 
Bin(-i)? 

2.  What  is  the  value  of  arc  tan  1?    Of  arc  cos  0?    Of  arc 
sec  3? 

3.  What  is  the  value  of  arc  sin  \  +  arc  cos  |  ?     Of  arc  tan 
1  +  arc  cot  1  ? 

4.  What    is    the    value    of    sin  (arc  sin  |)  ?    Of    cos  (arc 
sin  I)? 

5.  What    is    the    value    of    sin  (arc  sin  \  -f-  arc  sin  \  V§)  ? 
By  the  addition  theorem  this  may  be  expressed  as  sin  (A  +  B) 
=  sin  A  cos  B  +  cos  A  sin  B,    where  A  =  arc  sin  \    and   B  = 
arc  sin  \  A/3.     Then 

sin  (arc  sin  (|  •  \  +  \  V3  •  \  V3))  =  sin  (arc  sin  1)  =  1. 

6.  If  x  =  sin  A  and  y  =  sin  B,  show  that  (A  +  5)  =  arc 
sin  (x  Vl  -yz  +  y  Vl  -a;2). 

7.  Find  x  from  arc  sin  0.5  =  x. 


GENERAL  DIRECTIONS  91 

8.  Find  x  from  arc  tan  15  =  x. 

9.  Solve  for  x  in  the  equation  arc  sin  x  =  arc  cos  (x  —  ?). 
Note.  —  Take  the  sine  of  both  sides  first. 

10.  Solve  for  x  in  sin  (x  -  25°)  =  0.6. 

Note.  —  Expand  the  left  side  by  the  addition  theorem,  and 
use  48. 

11.  Solve  for  x  in  arc  tan  x  =  arc  sec  x  —  45°. 

12.  Solve  for  x  in  arc  tan  x  +  arc  cot  x  =  90°. 

13.  Solve  for  x  in  tan  (arc  tan  x)  =  1. 

55.  One  of  the  chief  uses 
of  the  trigonometric  functions 
is  found   in    the   solution  of 
problems  relating  to  triangles. 
All  questions  relating  to  right 
triangles  can  be  solved  by  di- 
rect use  of  the  definitions  of 

go 

47,  for  the  first  quadrant.    For  FIG  3g 

this  purpose  the  definitions  for 

acute  angles  can  be  modified  as  follows:  Consider  the  A  ABC 

in  the  figure,  C  being  the  right  angle.     Then 

y      side  opposite  A 
smA  =  ~  =  -  —, 

h         hypotenuse 

x      side  adjacent  A 

cos  A  =  T  =  — j—  » 

h         hypotenuse 

y      side  opposite  A 
tan  A  =  £  =  -  -— j~ 

x      side  adjacent  A 

Similar  relations  hold  for  the  angle  B. 

56.  General  directions  for  solving  problems  relating  to 
triangles: 

1.  Make  a  fairly  accurate  freehand  diagram  from  the  given 
conditions. 

2.  Mark  all  known  measurements  on  the  diagram.    Note 
the  position  and  relations  of  the  unknowns. 

3.  Select  a  formula  which  will  contain  one  of  the  unknowns 
and  the  knowns. 


92 


THE  TRIGONOMETRIC  FUNCTIONS 


4.  Substitute  the  values  of  the  knowns  in  the  formula  and 
solve  the  resulting  equation  for  the  unknown. 

5.  When  no  formula  fits  the  case  directly,  designate  auxiliary 
unknowns   by   symbols.     Formulate   as   many   equations   as 
unknowns  and  solve  the  system  of  equations  simultaneously 
as  in  algebra. 

1.  Find  the  height  of  a  tree,  having  given  the  angle  of  ele- 
vation, A  =  35°,  and  the  distance  AB  =  125'. 


125' 

FIG.  37. 


Note.  —  The  given  measurements  are  the  angle  A  and  the 
side  adjacent.  The  desired  measurement  is  the  side  opposite 
A.  We  select,  therefore,  the  tangent  ratio.  Write 

CB 


AB 


Substituting  values 


From  the  table 


Solving  for  CB, 


tan  A  = 


tan  35°  = 


0.7002  = 


CJB 
125' 


CB  =  87.53. 


2.   From  the  ends  of  a  line  AB  perpendiculars  are  dropped 
on  MN  meeting  MN  in  C  and  D  respectively.    The  angle 


GENERAL  DIRECTIONS 


93 


between  MN  and  AB  is  47°  30'.    The  line  AB  is  565  units  long. 
Find  CD. 


Fia.  38. 

Note.  —  CD  is  called  the  projection  of  AB  on  MN.    The 
angle  AEB  is  called  the  angle  of  projection.     See  Geometry. 

3.  Find  the  projection  of  a  line  50'  long  on  a  line  at  an  angle 
of  60°  with  it. 

4.  What  are  the  projec- 
tions on  the  axes  of  a  line 
120'  long  which  is  inclined 
36°  with  the  x-axis?    The 
lengths  of  AB  and  CD  in 
the  diagram  are  wanted. 

5.  A  plane  surface  may     

be  projected  on  a  plane  in- 
clined to  it  in  a  manner 
exactly  similar  to  the  pro- 
jection of  a  line  on  another 

line. 

Find  the  area  of  the  projection  of  a  rectangle  30'  x  60'  on  a 
plane  inclined  75°  to  it. 

6.  A  roof  is  30'  x  20'  and  is  inclined  37°  to  the  horizontal. 
How  many  sq.  ft.  of  floor  does  it  cover? 

7.  What  is  the  area  of  the  projection  of  a  circle  of  10'  radius 
on  a  plane  inclined  to  it  at  an  angle  of  27°  ? 

8.  A  circle  has  a  radius  of  10'.    A  chord  of  the  circle  is  15' 
long.     How  far  is  the  chord  from  the  center? 

9.  Find  the  perimeter  of  a  regular  pentagon  inscribed  hi  a 
circle  of  radius  5". 


FIG.  39. 


94 


THE  TRIGONOMETRIC  FUNCTIONS 


10.   Find  BD  in  the  diagram  if  AC  =  100',  angle  A  =  25° 
and  angle  C  =  30°. 

D 


FIG.  40. 

Note.  —  Introduce  the  auxiliary  unknown  x  for  CB  and 
formulate  two  equations.     Then  eliminate  x. 

11.  What  is  the  eastward  component  of  a  force  of  105  Ibs. 

acting  in  a  direction  30°  east  of 
due  north?  The  length  OM  rep- 
resents the  eastward  component. 
Find  the  northward  component. 

12.  In  the  triangle  ABC,  A  = 
42°,  AB  =  125,  AC  =  150.    Find 

X    the  altitude  on  the  side  AC  and 

the  area  of  the  triangle.  From 
this  problem  determine  a  formula 
for  the  area  of  any  triangle  when 
two  sides  and  their  included  angle 
are  given. 

13.  By  drawing  an  equilateral 

triangle  and  its  altitude  derive  values  of  sin  30°,  cos  30°,  sin  60°, 
cos  60°. 

14.  By  considering  a  square  and  its  diagonal  derive  values  of 
sin  45°,  cos  45°. 

15.  A  travels  north  50  mi.,  then  37°  east  of  north  a  distance 
75  mi.,  then  10°  west  of  south  125  mi.    Find  the  length  and 
direction  of  the  line  from  the  starting  point  to  his  final  position. 


FIG.  41. 


THE  SINE  LAW  OF  TRIANGLES 


95 


Find  the  distance  east  or  west  he  traveled  and  the  distance 
north  or  south,  from  the  starting  point. 
67.  The  sine  law  of  triangles.  —  From  the  figure  it  is  seen 

that 

h  =  c  sin  A  and  h  =  a  sin  C. 

.'.     c  sin  A  =  a  sin  C 
and 


sin  C      sin  A 


FIG.  42. 

By  drawing  altitudes  from  the  other  vertices  the  equations 

b  c 


and 


sin  B      sin  C 
a  b 


sin  A      sin  B 
can  be  derived  in  exactly  the  same  way  as  above. 
a  b  c 


(39) 


sin  A      sin  B      sin  C 


These  equations  are  known  as  the  sine  law.  Its  use  is  indi- 
cated when  a  side  and  the  angle  opposite  are  among  the  parts 
to  be  considered. 

68.  The  cosine  law  of  triangles.  —  From  the  figure  of  the 
last  section, 

c*  =  w  +  s2  =  h2  +  (b  -  t/)2  =  h2  +  62  -  2by  +  y\ 


96 


THE  TRIGONOMETRIC  FUNCTIONS 


But  h  =  a  sin  C,    y  =  a  cos  C.      Substituting    in    the    above 
equation 

c2  =  62  -  2abcosC  +  a£sin2C  +  a2cos2C 
(40)      or     c2  =  a2  +  b2  -  2  ab  cos  C. 

By  using  the  other  sides  of  the  triangle  as  bases  in  turn,  the 
following  are  derived  in  the  same  way: 
(40a)  a2  =  b2  +  c2  -  2  be  cos  A. 

(406)  b2  =  a2  +  c2  -  2  ac  cos  B. 

59.  Example  of  the  use  of  the  sine  law.  —  Given  A  ••=•• 
15°  19',  C  =  72°  44',  c  =  250.4,  of  the  triangle  ABC,  to  find 
the  remaining  parts.  Immediately 


B  =  18G-(A 


)  =  91°  57'. 
By  the  sine  law, 


250.4 


_ 

sin  15°  19'      sin  72°  44' 
250.4  sin  15°  19' 


sin  72°  44' 

In  logarithms  this  is 
log  a  =  log  250.4  +  tog  sin  15°  19'  -  log  sin  72°  44' 
=       2.3986 
+  9.4218-10 
11.8204-10 
-  9.9800-10 

1.8404 
/.     a  =  69.  25. 

Again,  to  find  b,  using  the  sine  law, 

sin  72°  44'      sin  91°  57' 


Solving  as  above, 


250.4  b 

6  =  262.0. 


1.  Student  check  the  result  with  natural  functions,  using  the 
slide  rule. 


EXAMPLE  OF  THE  USE  OF  THE  COSINE  LAW 


97 


2.  Construct  the  triangle  to  scale  50  to  1"  from  the  given 
data  and  then  measure  the  unknown  parts.  Check  with  the 
calculated  values. 

60.  Example  of  the  use  of  the  cosine  law.  —  Given  a  = 
1686,  6  =  960,  C  =  128°  04',  to  find  the  other  parts. 

By  the  cosine  law 

c2  =  16862  +  9602  -  2  •  1686  •  960  -  cos  128°  4'.* 
Whence  c  =  2400. 

The  values  of  A  and  B  may  now  be  found  by  the  sine  law. 


1.  Find  AB,  the  distance  across  a  river,  from  the  data  given 
in  the  diagram.  C  is  a  point  on  top  of  a  hill,  AB  and  CD  are 
horizontal  lines.  BC  =  500',  angle  DC  A  =  15°,  angle  CBE  = 
20°. 


River 


FIG.  45. 


2.  Two  trains  leave  a  station  at  the  same  time  on  straight 
tracks  inclined  to  each  other  at  an  angle  of  35°.  Train  A  travels 
25  mi.  per  hr.,  train  B  travels  35  mi.  per  hr.  How  far  apart  are 
the  trains  at  the  end  of  3  hrs.  ? 

*  Note  that  cos  128°  4'  is  negative. 


98  THE  TRIGONOMETRIC  FUNCTIONS 

3.  Find  A B  from  the  measurements  given  below.   CD  =  1000', 
ACD  =  120°,  DCB  =  30°,  ADC  =  33°,  CDB  =  105°. 

4.  Use  the  cosine  law  to  find  the  angles  of  the  triangle  ABC, 
if  a  =  75.8,  6  =  64.2,  c  =  81.9. 

5.  A  tower  stands  at  the  top  of  a  slope  inclined  20°  with  the 
horizontal.     From  a  point  800'  down  the  slope  from  the  tower 
the  angle  subtended  by  the  tower  is  5°  25'.     Find  the  height 
of  the  tower. 

6.  The  tripod  of  a  camera  stands  on  a  hillside.     One  leg  is 
3'  long,  the  other  two  legs  are  each  5'  and  set  on  the  ground  at 
the  same  level.     The  three  legs  make  equal  angles  with  each 
other  successively  around.    These  angles  are  each  38°.     Find 
the  distance  between  the  feet  of  the  legs. 

61.  Conversion  formulas.  —  The  cosine  law  does  not 
admit  of  use  with  logarithms.  For  this  reason  another  law, 
derived  from  the  sine  law,  is  used  when  it  is  desired  to  employ 
logarithmic  calculations.  This  law  is  known  as  the  tangent 
law.  Before  the  tangent  law  can  be  given  some  formulas  must 
be  developed. 

From  the  addition  theorems  we  have: 

1.  sin  (A  +  B)  —  sin  A  cos  B  +  cos  .A  sinB. 

2.  sin  (A  —  B)  =  sin  A  cos  B  —  cos  A  sin  B. 

3.  cos  (A  -f  J5)  =  cos  A  cos  B  —  sin  A  sin  B. 

4.  cos  (A  —  B)  =  cos  A  cos  B  +  sin  A  sin  B. 

Add  (2)  to  (1);  subtract  (2)  from  (1);  add  (4)  to  (3);  subtract 
(4)  from  (3),  and  obtain  the  following  equations  in  order. 

5.  sin  (A  +  B)  +  sin  (A  -  B)  =  2  sin  A  cosB. 

6.  sin  (A  +  5)  -  sin  (A  -  B)  =  2  cos  A  sinB. 

7.  cos  (A  +  B)  +  cos  (A  -B}  =  2  cos  A  cos  B. 

8.  cos  (A  +  B)  -  cos  (A  -  &)  =  -2  sin  A  sinB. 

P  +  Q 
In    the    last    four    equations    substitute    A  =  — » —    and 

•  •  •* ' 

B  =      7  ^.    There  results: 


THE  TANGENT  LAW  99 

(41)  sin  P  +  sin  Q  =  2  sinP^^cosP  ~  ^  • 

a  & 

P  -\-  O       P  —  O 

(42)  sinP-sinQ  =  2cos     ^vsin      g  v. 

P  4-  O       P  —  O 

(43)  cosP  +  cos(>  =  2cos— jp-cos — —£• 

a  a 

P  4-O       P  —  O 

(44)  cos  P  -  cos  Q  =  -2  sin      ;.      sin      » 

62.   The  tangent  law.  —  By  the  sine  law 
a  _  sin  A 
b      sinB 

Taking  this  proportion  by  composition  and  division  gives 

a  4-  b  _  sin  A  -j-  sin  B 
a  —  b      sin  A  —  sin  B 

Applying  (41),  (42),  to  the  right  side  gives,  by  use  of  48, 

fAK\  a  +  6 

(45) r  = 


Two  similar  equations  using  the  other  parts  of  the  triangle  are 
obtained  in  a  similar  manner.  This  equation  is  called  the 
tangent  law  for  triangles.  This  law  applies  directly  to  the 
case  that  was  solved  by  the  cosine  law  in  60.  That  problem 
will  now  be  solved  by  the  tangent  law.  Write 

1686  +  960      tan  \  (180°  -  128°  04') 

1686  -  960  "         tan  \(A  -  B) 

since  \  (A  +  5)  =  \  (180°  —  C)  in  any  triangle.  Solving  the 
above  equation  for  tan  \  (A  —  B), 

726  •  tan  25°  58' 


Applying  logarithms  to  this  equation, 

log  tan  i  (A  -  5)  =  log  726  +  log  tan  25°  58'  -  log  2646. 
Substituting  the  values 

log  tan  HA  -  B)  =  9.1258  -  10. 
Whence  |  (A  -  B}  =  7°  37'. 


100  THE  TRIGONOMETRIC  FUNCTIONS 

Now    %(A  +  B)+%(A-B)  =  A  =  25°  58'  +  7°  37'  =  33°  35', 
and     %(A+B)  -%(A-B)  =  B  =  25°  58'  -  7°  37'  =  18°  21'. 

Having  the  angles  A  and  B,  the  remaining  side  can  be  found 
by  the  sine  law. 

63.  Ambiguous  case.  —  When  the  given  parts  of  a  triangle 

are  two  sides  and  an  angle  op- 
posite one  of  these  sides  any 
one  of  the  following  results  may 
occur: 

Let  a,  c,  A  be  the  given  parts 
and  let  h  be  the  altitude  from  B. 
1.   a  <  h,  where  h  =  c  sin  A. 
Under  these  circumstances  the 
triangle  cannot  be  constructed. 

2.  a  =  h,  where  h  =  csin  A.    There  is,  under  these  circum- 
stances, one  triangle,  a  right  triangle  with  the  right  angle  at  D. 

3.  a  >  h  and  a  <  c,  where  h  =  csin  A.     Under  these  con- 
ditions there  will  be  two  triangles,  ABC  and  ABC'. 

Note.  —  C'  +  C  =  180°.  Hence  sin  C"  =  sin  C.  In  solving 
this  case  the  value  of  C  will  be  obtained  from  its  sine.  Since 
there  are  two  angles  each  less  than  180°  having  the  same  sine,  a 
choice  must  be  made  between  C'  and  C.  This  choice  must  be 
based  on  other  data  in  the  problem.  When  no  conditions  are 
given  for  making  a  choice  both  solutions  must  be  given. 

Note.  —  Whenever  in  solving  a  problem  a  result  is  obtained 
that  implies  that  the  sine  or  cosine  of  an  angle  is  greater  than  1, 
the  problem  is  either  impossible  or  an  error  has  been  made  in  the 
calculations. 

(a)  Example.  —  Given  a  =  250,  A  =  42°  12',  c  =  600,  to 
find  the  remaining  parts  of  the  triangle  ABC.  By  the  sine  law, 

sin  C      sin  42°  12' 


600  250 

Solving  gives 

log  sin  C  =  10.2075  -10. 


DOUBLE  AND  HALF-ANGLE  FORMULAS  101 

This  is  equivalent  to  sin  C  >  1,  which  is  impossible.     This 
means  250  <  h  and  the  triangle  cannot  be  constructed. 

(b)  Example.  —  Given  c  =  254.3,   a  =  396.8,   A  =  94°  29', 
to  find  the  remaining  parts  of  the  triangle  ABC. 

By  the  sine  law 

396.8      =  254.3 
sin  94°  29'      sinC' 
Solving, 

log  sin  C  =  9.8054  -10. 
/.     C  =  39°  43' or  140°  17'. 

Since  an  obtuse  angle  occurred  in  the  given  data,  only  the 
smaller  value  of  C  can  be  used. 

(c)  Example.  —  Given  a  =  250,   c  =  300,   A  =  42°  12',   to 
find  the  remaining  parts  of  the  triangle  ABC.    By  the  sine  law, 

sin  C  =  sin  42°  12' 
300  =         250 
Solving, 

log  sin  C  =  9.9064  -10. 
.-.     C  =  53°  43'  or  126°  17'. 


K b *i 


In  this  case  either  value  of  C  will  satisfy  the  problem  and  there 
are  two  possible  triangles.  The  other  parts  of  each  triangle 
can  be  found. 

64.  Double  and  half-angle  formulas.  —  In  the  addition 
theorems  put  B  =  A.  Then 

(46)          sin  (A  +  A)  =  sin  2  A  =  2  sin  A  cos  A. 


102  THE  TRIGONOMETRIC  FUNCTIONS 

47.  cos  (A  +  A)  =  cos  2  A  =  cos2  A  -  sin2  ,4  =  2  cos2  A  -  1 


From  47,  2  sin2  A  =  1  -  cos  2  4 


,..  .  .  /I  —  cos 2 A 

or    (1)  sin  A  =  y  -       - 

From  47  again,  2  cos2  A  =  1  +  cos  2  A 
or    (2)  cos  A  = 


In  (1)  and  (2)  put          A  =  -%  and  obtain: 
(48)  sinlp: 


(49)  cos 


1  /1  +  cosP 

2r      V          2 


As  an  exercise  derive  tan  \  P,  cot  \  P,  sec  \  P,  esc  |  P. 

65.  Angles  of  a  triangle  in  terms  of  its  sides.  —  By  the 
cosine  law: 

52    •     g2  _  a2 

(1) 

Add  1  to  both  sides, 

62 
(2)   l+cosA=- 


Subtract  both  sides  of  (1)  from  1, 

a2  —  (b2  —  2 be  +  c2)  _  (a-j-b  —  c)  (a— b+c) 

Substituting  these  values  in  (48),  (49),  64,  gives: 

(K.f]\     eini  A    — 

Wl   sin 75/1  — 


(51)  cosl, 


RADIUS  OF  CIRCLE 


103 


where  a  +  6  +  c  =  2s.    Note  the  arrangement  of  letters  in 
the  last  two  equations. 
A  is  any  angle  of  the  triangle: 

1 .  Derive  tan  ~  A  from  (48) ,  (49) .  Tan  ~  A  =  V r-^ — r — -  • 

z  &  §  \§  —  a) 

2.  Find  the  angles  of  the  triangle  ABC  it  a  =  45,  6  =  55, 
c  =  66. 

66.   Radius   of   circle  inscribed   in  a  triangle  ABC.    Call 
a  +  6  +  c  =  2s.     By  the  geometry  of  the  figure: 

A2  +  53  +  Cl  =  s 

or  A2  =  s  -  £3  -  Cl  =  s  -  53  -  C3 

=  s  —  a. 


Now 
or 

(52) 


r  =  (s—  a) 


where  a  is  any  side  and  A  the  opposite  angle  of  the  triangle. 

67.   Radius  of  circle  circumscribed  about  a  triangle  ABC. 
In  Fig.  49, 

angle  A'  =  angle  A, 
angle  A  BC  =  90°, 


104 

Therefore 

and 

(53) 


THE  TRIGONOMETRIC  FUNCTIONS 


.     ,        .    „,      BC      BC       a 
sin  A  =  sin  A'  =  -TT7l  =  ^-5  = 
' 


-TTl       -5        -5 
A'C     2R      2R 


a 


^  -r 
2  sin  A 

where  A  is  any  angle  of  the  triangle  and  a  the  opposite  side. 


68.   Circular  measure  of  angles.  —  We  are  accustomed  to 
measure  angles  in  degrees.     It  is  often  convenient  to  measure 
angles  with  another  unit.     This  unit  is 
called  the  radian  and  is  denned  by  the 
equation, 


(54) 


BicAB  =  r  -9, 


where  r  is  the  radius  of  the  arc  AB  and  0 
is  the  angle  in  radians.  Since  an  arc 
equal  to  the  circumference  has  the  same 
measure  as  a  360°  angle  at  the  center  the 
relation  between  the  degree  and  the  radian  is  easily  obtained. 


FIG.  50. 


CIRCULAR  MEASURE  OF  ANGLES 


105 


By  (54),  9  =  —  in  radians,  that  is,  the  arc  divided  by  its 
radius  gives  the  measure  of  the  subtended  angle  in  radians. 

Therefore 
(55)     and 


arc   (circumference)  _  2irr  _ 
r  r 


2  TT  radians  =  360°. 
360 


1  radian  = 


27T 


=  57°  17'  45"- 


The  following  table  of  equivalents  is  easily  derived  and  will 
be  used  in  work  to  follow.  It  would  be  well  to  memorize  this 
table. 


Radians. 

Degrees. 

2x 

360 

IT 

180 

7T/2 

90 

7T/3 

60 

r/4 

45 

1T/6 

30 

1.   The  radius  of  a  circle  is  12.     The  angle  at  the  center  is 

7T 

45°  =     radians.     Find  the  arc  intercepted  on  the  circumference. 


By  (54), 


arc=  12- =  ST. 


2.  Find  the  arc  subtended  by  a  chord  4'  from  the  center  of 
a  circle  of  radius  10'. 

The  cosine  of  half  the  angle  subtended  by  the  chord  is  T%  = 
0.4.  From  this  the  angle  is  found  from  the  tables  to  be  23°  35' 
or  0.41+  radians.  Now  by  (52), 

arc  =  10  X  .41+  =  4.1+,  ft. 

3.  In  a  circle  of  radius  12"  a  line  10"  from  the  center  is 
drawn.    What  is  the  length  of  the  arc  cut  off?     (First  find  the 
angle  subtended  by  the  chord,  then  apply  Eq.  (52.)) 


106  THE  TRIGONOMETRIC  FUNCTIONS 

4.  The  radius  of  a  circle  is  15,  a  chord  is  drawn  cutting  off 
a  segment  of  altitude  3.     Find  the  area  of  the  segment. 

5.  A  carriage  wheel  is  4'  in  diameter,  the  carriage  is  traveling 
10  mi.  per  hr.    Find  the  number  of  revolutions  per  min.  and 
the  number  of  radians  per  sec.  turned  through  by  the  wheel. 

6.  The  arc  through  which  a  pendulum  swings  is  4".    The 
length  of  the  pendulum  is  39.4".     Find  the  angle  of  swing  in 
radians  and  in  degrees. 

7.  An  arc  is  10"  and  the  angle  measured  at  the  center  is  1| 
radians.     Find  the  radius  of  the  circle. 

8.  A  horizontal  tank  6'  in  diameter  and  30'  long  is  filled  to  a 
depth  of  2'.     Find  the  number  of  gallons  in  the  tank.     See 
Chap.  II  (51). 

68a.   Mil,   a  unit  of  angular  measure:  Just  as  a  central 
angle  standing  on  an  arc  that  is  -5^  of  the  circumference  of  a 
circle  is  one  degree,  so  also  a  central  angle  standing  on  an  arc 
that  is  s5Vrr  of  the  circumference  is  one  mil.    Since  by  this 
definition  there  are  6400  mils  at  the  center  of  a  circle,  it  follows 
that  the  length  of  the  arc  that  subtends  one  mil  is 
circumference  _  2irr  _  6.283 r  _  mftQQ1  ~ 
6400         ~  6400  ~  "6400"  : 

1.  In  military  work  it  is  common  to  speak  of  a  mil  as  ap- 
proximately equal  to  an  arc  of  one  foot  at  a  distance  of  1000 
feet  or  one  yard  at  a  distance  of  1000  yards.     Give  reason  for 
this  statement. 

2.  In  a  table  of  natural  sines  of  angles,  expressed  in  mils,  is 
the  following: 

Angle  in  mils        64  1600        2400         3180 

sine        .0628  1         .6071         .0194 

Check  this  table  by  converting  mils  to  degrees  and  then  use 
table  of  natural  sines. 

3.  From  problem  2  find  the  sine  of  800  mils. 

69.  Explanatory  definitions.  —  (a)  In  surveying  the  direc- 
tion of  a  line  is  usually  given  by  giving  the  angle  which  it  makes 
with  a  north  and  south  line.  If  the  line  runs  north  and  east  its 


PROBLEMS 


107 


direction  is  designated  by  N.  6°  E.,  where  6°  is  the  angle  between 
a  line  running  due  north  from  the  starting  point  and  the  line  of 
sight  from  same  point.  If  the  line  runs  in  a  northwesterly 
direction  its  direction  is 
designated  by  N.  6°  W. 
Similar  notations  apply 
to  directions  southeast- 
erly and  southwesterly. 

(6)  The  angle  of  ele- 
vation of  a  point  is  the 
angle  between  a  horizon- 
tal line  and  a  line  from 
the  observer  to  the  ob- 
served point,  above  the 
level  of  the  observer. 

(c)  The  angle  of  de- 
pression of  a  point  below 
the  observer  is  the  angle 
between  a  horizontal  line 
and  a  line  from  the  observer  to  the  point  observed.  In  the 
figure,  9  is  the  angle  of  elevation  of  B  from  A,  and  tf  is  the 
angle  of  depression  of  C  from  A. 

PROBLEMS. 

1.  The  distance  between  two  points  measured  on  a  slope  of  5°  42'  with 
the  horizontal  is  210.3'.     Find  the  horizontal  distance  between  the  points. 

2.  The  distance  between  the  points  A  and  B  measured  on  the  horizontal 
is  388.0'.    The  bearing  from  A  to  B  is  N.  30°  E.    How  far  is  B  north 
and  east  of  A? 

3.  A  surveyor  sets  his  instrument  over  a  stake  at  A  and  reads  the 
bearing  to  another  stake  at  B,  S.  22°  10'  E.     The  distance  from  A  to  B  is 
1142.1'  measured  on  an  upward  slope  of  12°  21'.    Find  how  far  B  is  north 
or  south  of  A.     How  far  is  B  east  or  west  of  A?    How  far  is  B  above  or 
below  A?    (Save  results.) 

4.  Compute  similarly  the  position  of  C  with  reference  to  B  when  the 
bearing  from  B  to  C  is  S.  37°  30'  W.  and  the  distance  843.7'  measured  on 
a  downward  slope  of  10°  25'.     (Save  results.) 

5.  Using  the  results  of  (3)  and  (4)  compute  the  position  of  C  with 
respect  to  A. 


FIG.  51. 


108  THE  TRIGONOMETRIC  FUNCTIONS 

6.  It  is  desired  to  drive  a  straight  passage  from  A  to  B  in  a  mine.    A  is 
known  to  be  3500'  north  and  2200'  west  of  a  third  point  C.     B  is  known  to 
be  2500'  south  and  600'  west  of  C.    Find  the  length  and  bearing  of  the 
straight  passage  from  A  to  B. 

7.  A  horizontal  line  1033  ft,  is  measured  in  the  same  vertical  plane 
with  the  top  of  a  mountain.     From  one  end  of  the  line  the  angle  of  elevation 
of  the  top  of  the  mountain  is  13°  22',  from  the  other  end  the  angle  of  ele- 
vation is  5°  10'.     What  is  the  height  of  the  mountain  above  the  horizontal 
line? 

8.  On  the  right  bank  of  a  river,  two  stakes  A  and  B  are  set  250.0' 
apart  on  a  horizontal  line.     On  the  left  bank  a  stake  C  is  set  so  that  the 
angle  A  in  the  triangle  ABC  is  90°.    The  angle  B  is  measured  to  be  38°  41'. 
What  is  the  distance  from  A  to  C  ? 

9.  From  a  hill  top  the  angles  of  two  points  on  opposite  sides  of  the  hill 
are  20°  33'  and  15°  10',  respectively.    The  distance  from  the  hill  top  to 
the  first  point  is  1200  yds.     The  distance  to  the  second  point  2000  yds 
Find  the  distance  between  the  points. 

10.  A   surveyor  took   measurements   as   follows:    AB  =  500',   angle 
DAB  =  100°,  angle  DAC  =  67°  45',  angle  ABC  =  125°,  angle  DBG  = 
70°.    Find  the  length  of  DC. 

11.  A  flag  pole  stands  on  a  tower  50' high.     From  A,  on  a  level  with  the 
base  of  the  tower,  the  angle  of  elevation  of  the  top  of  the  tower  is  42°  35'. 
From  A  the  pole  subtends  an  angle  of  10°  15'.     Find  the  length  of  the  flag 
pole. 

12.  Find  the  difference  between  the  areas  of  the  triangle,  ABC,  and  its 
inscribed  and  circumscribed  circles,  if  the  sides  of  the  triangle  are  a  =  56.3, 
b  =  76.5,  c  =  68.8. 

13.  What  are  the  angles  of  the  triangle  whose  sides  are,  a  =  665,  6  = 
776,  c  =  887? 

14.  Two  stones  are  one  mile  apart  on  a  straight  road,     From  a  tower 
standing  on  the  roadside  the  angles  of  depression  of  the  stones  are  15°  and 
3°  50',  respectively.     Find  the  height  of  the  tower  and  its  distance  from 
each  stone. 

Note.  —  There  are  two  cases:   (a)  when  the  tower  is  between  the  stones; 
(6)  when  the  stones  are  on  the  same  side  of  the  tower. 

70.   Graphic  representation  of  the  trigonometric  functions. — 

(a)   Construct  the  graph  of  the  equation 

y  =  sin  x. 

To  do  this,  the  values  of  the  angle  x  must  be  laid  off  in  linear 
units  to  some  scale.  From  68,  if  r  =  1,  the  arc  =  the  angle  at 
the  center  expressed  in  radians.  This  furnishes  the  unit  of 


GRAPHIC  REPRESENTATION 


109 


measure,  viz.,  an  arc  equal  in  length  to  the  radius.     Lay  off 
values  of  x  along  the  re-axis  to  the  scale  of  1  radian  =  1". 
Tabulate  values  of  x,  and  y  =  sin  x  below. 


z 

y  =  sin  x. 

X 

y  =  sin  i. 

0°  =  0       radians*.. 

0.0 

210°  =  7  ir/6  radians.. 

-0.5 

30°  =  7T/6 

0.5 

225°  =  5x/4 

-0.707 

45°  =  x/4 

0.707 

240°  =  47T/3 

-0.866 

60°  =  x/3 

0.866 

270°  =  3T/2 

-1.0 

90°  =  7T/2 

1.0 

300°  =  5x/3 

-0.866 

120°  =  27T/3 

0.866 

315°  =  7  7T/4 

-0.707 

135°  =  3  x/4 

0.707 

330°  =  llT/6 

-0.5 

150°  =  5  ir/6 

0.5 

360°  =  2  T 

0.0 

180°    =   7T 

0.0 

,2!      2L 
6       |3 

<-r=l-H 


137T 


FIG.  52. 

A  property  of  trigonometric  functions  known  as  periodicity 
is  illustrated  by  this  curve.  This  was  suggested  in  52.  It  is 
noticed  that  as  a  point  traces  the  above  curve  the  values  of  y 
are  the  same  for  positions  of  the  point  going  in  the  same  direc- 
tion on  the  curve,  such  as  the  pairs  of  points  A,  B;  A',  B';  A", 
B".  It  follows  that  corresponding  to  any  value  of  sin  x,  there 
are  infinitely  many  values  of  x  differing  successively  by  360°  or 
2  TT  radians.  This  fact  may  be  expressed  by 

sin  x  =  sin  (x  ±  n  360°)  =  sin  (x  ±  2  rnr), 


*  Note  that  T  =  3.1416,  then  ir/6  radians  =  0.5236  radians  and  similarly 
with  other  values. 


110  THE  TRIGONOMETRIC  FUNCTIONS 

where  n  =  0,  1,  2,  3,  ...     A  similar  relation  holds  for  each 
of  the  functions  cos  x,  tan  x,  cot  x,  sec  x,  esc  x. 

It  may  be  further  noted  that  there  are  two  equal  ordinates 
of  the  curve  y  =  sin  x  in  each  of  the  intervals  0°  to  180°;  180° 
to  360°;  360°  to  540°,  etc.  From  this  fact  it  occurs  that  the 
value  of  a  single  function  alone,  say  of  sin  x  or  tan  x,  is  not  suffi- 
cient to  determine  which  of  the  two  values  of  x  between  0°  and 
180°  or  between  180°  and  360°  is  to  be  taken.  Either  another 
function  must  be  known  or  else  the  quadrant  in  which  the  angle 
lies  must  be  known.  This  fact  was  illustrated  in  the  solution 
of  the  ambiguous  case  of  triangles,  63. 

1.  Construct  the  graph  of  y  =  cos  x  in  a  manner  similar  to 
that  employed  in  constructing  the  graph  of  y  =  sin  x,  above,    j 

2.  Construct  the  grapk  of  y  =  sin  x  +  cos  x. 

Note.  —  Tabulate  values  of  sin  x  and  cos  x  and  add  the  corre- 
sponding values.  Use  these  sums  as  ordinates  of  the  curve. 

3.  Construct  the  graph  of  y  =  sin  2  x. 

Note.  —  Erect  the  ordinates  equal  to  sin  2  x,  over  the  values 
of  x. 

4.  In  a  manner  similar  to  that  indicated  in  Ex.  (3),  construct 
the  graph  of  y  =  sin  3  x.     How  do  the  curves  in  this  exercise 
and  the  preceding  differ  from  the  graph  of  y  =  sin  x  ? 

5.  Construct  the  graph  of  y  =  sin  ^  x. 

6.  Construct  the  graph  of  y  =  tan  x.    What  peculiarity  has 
this  curve  at  x  =  ir/2,  x  =  3  ir/2,  etc.? 

7.  Construct  the  graph  of  y  =  -  sin  x,  given  that  the  value 

dp 

,.  r    ., /sinx\ 

of  limit 1  =  1. 

*-»o  \  x  I 

8.  Construct   the   graph   of   y  =  sec  x.     What   peculiarity 
does  this  curve  have  at  ir/2,  3  7r/2,  etc.  ?     In  what  way  does  it 
differ  from  the  sine  curve  and  the  tangent  curve  at  x  =  0, 
x  =  TT,  etc.  ? 

71.  The  equation  y  =  arc  sin  x  may  also  be  studied  by  means 
of  its  graph.  This  function  is  infinitely  many  valued  as  was 
shown  in  52. 


GRAPHIC  REPRESENTATION 


111 


Values  of  y  =  arc  sin  x  for  different  values  of  x. 


0 


0.5 


0.866 


-1.0 


-0.5 


-0.866 


-1.0 


±360°,  ±720°,  .  .  . 

0° 

±27T,  ±47T,    .    .    . 

30°, 

30°  ±360°,  .  .  . 

•     7T 

7r        o 

6' 

g±27T,      ... 

150°, 

150°  ±  360°,  ...                     ^ 

*-—/B 

STT 

STT                                                   137T 

^/ 

T' 

6  ±2»,  ...                         -g- 

-/B 

60°, 

60°  ±360°,  ...                        ,/ 

117T 

•  if 

""           O                                                                         f~* 

6 

3' 

3  ^  27r>  •  •  •                       /_ 

22L 

3 

120°, 

120°  ±  360°,  ...             / 

3T 

2  IT      " 

2                                            \ 

2 

—  :r-» 

±  2?r    .                            V 

47T 

3 

3                                           \T 

3 

90°, 

90°  ±360°,  .  .  .                   \-- 

~6~ 

•7T2 

2' 

I*!..,.:,                                          > 

\ 

5ir 

210°, 

210°  ±360°,  ...                       6 

.                                                                   2ir 

\ 

7T 

i  7T 

A 

T1 

O                                                                               "" 

\ 

'2 

~~j 

330°, 

330°±360°,  ...                       ^ 

jr.' 

UTT 

[     O  _.                                                                       —• 

6   ' 

6            '»»••<                          r-i  T 

/ 

oAft° 

Z4U  , 

47T 

4x                                            AV~- 

7T 

M-i     „ 

3  ' 

3  :b2T,  .  .  .                       ^T1 

6 

300°, 

300°  ±  360°,  ...                      Fie 

i.  53. 

270°, 

270°  ±360°,  .  .  . 

37T 

37T 

T 

2   ±Zir,  .  .  . 

7T  /  TT 

The  figure  shows  the  graph  for  values  of  y  from  —  ^  to  +  -5-- 

o  o 


112  THE  TRIGONOMETRIC  FUNCTIONS 

The  curve  shows  that  any  line  parallel  to  the  y-axis  and  less 
than  a  unit  distant  from  it  cuts  the  curve  in  many  points, 
that  is,  there  are  many  ordinates  for  each  abscissa.  Such  a 
function  as  y  =  arc  sin  x  is  called  a  multivalued  function  of  x. 
The  smallest  positive  ordinate  for  any  value  of  the  abscissa  is 
called  the  principal  value  of  the  ordinate  or  angle. 

1.  Construct  the  graph  of  y  =  arc  cosx. 

2.  Construct  the  graph  of  y  =  arc  tan  x. 

3.  Construct  the  graph  of  y  =  arc  sin  2  x. 

4.  Construct  the  graph  of  y  =  arc  sin  J  x. 

72.  Equations  involving  trigonometric  functions  as  unknowns 
are  of  frequent  occurrence.  They  may  be  solved  in  much  the 
same  way  as  algebraic  equations. 

1.  Find  the  principal  value  of  x  that  will  satisfy  the  equation 

2  sin2  x  —  3  cos  x  =  0. 
This  can  be  written, 

2  (1  —  cos2  x)  —  3  cos  x  =  0. 
Solving  for  cos  x, 

cos  x  =  0.5    or    2. 
/.     x  =  arc  cos  0.5     or    arc  cos  2. 

The  first  gives  x  =  60°  and  300°,  of  which  60°  is  the  principal 
value.  The  second  is  impossible  since  |  cos  x  \  =  1.  This  value 
must,  therefore,  be  rejected. 

2.  Solve  and  determine  the  principal  value  of  x  in 

2  sin  2  z  —  cos2  x  =  0. 

Note.  —  First  remove  the  double  angle  by  substituting  its 
equivalent  from  formulas  of  53. 

3.  Solve  cos2  e  —  sin  6  =  %. 

4.  "  4  cos  0  =  3  sec  x. 

5.  "  3sin0-2cos20  =  0. 

6.  "  tan  6  +  cot  0  =  2. 

7.  "  4sec0-7tan20  =  3. 


SPECIAL  INTERPOLATION 


113 


Note.  —  It  may  happen  in  solving  an  equation  containing 
two  functions,  that  eliminating  one  function  will  lead  to  only 
part  of  the  solution.  The  other  part  of  the  solution  may  be 
obtained  by  eliminating  the  other  function. 

72a.  —  Tables  requiring  special  interpolation.  —  Gradients 
are  commonly  called  grades  or  slopes  and  are  expressed: 

(a)  By  the  angle  which  the  line  of  direction  makes  with  a 
horizontal  line. 

Example.  —  A  gradient  of  1°,  2°,  3°,  etc. 

(6)  By  the  change  of  elevation  corresponding  to  a  given 
horizontal  distance. 

Example.  —  An  elevation  of  4'  in  100',  that  is,  a  4  per  cent 
slope. 

An  elevation  1'  in  60'  gives  a  gradient  of  ^V  (read:  1  on  60). 

Below  is  a  table  which  gives  differences  of  elevation  for 
gradients  of  0°  to  5°,  and  horizontal  distances. 


Gradient  in 
degrees. 

Difference  of  elevation  for  horizontal  distance  of: 

1 

2 

3 

4 

5 

6 

7 

8 

9 

0.5 

1 
2 
3 
4 
5 

0.0087 
0.0175 
0.0349 
0.0524 
0.0699 
0.0875 

0.0174 
0.0350 
0.0698 
0.1048 
0.1398 
0.1750 

0.0261 
0.0525 
0.1047 
0.1572 
0.2097 
0.2625 

0.0348 
0.0700 
0.1396 
0.2096 
0.2796 
0.3500 

0.0435 
0.0875 
0.1745 
0.2620 
0.3495 
0.4375 

0.0522 
0.1050 
0.2094 
0.3144 
0.4194 
0.5250 

0.0609 
0.1225 
0.2443 
0.3668 
0.4893 
0.6125 

0.0696 
0.1400 
0.2792 
0.4192 
0.5592 
0.7000 

0.0783 
0.1575 
0.3141 
0.4716 
0.6291 
0.7875 

Find  the  difference  of  elevation  for  a  gradient  of  3°,  and  a 
horizontal  distance  of  567  ft. 
From  table,  on  line  with  gradient  of  3°, 

For  distance  7  the  elevation  is 0.3668 

For  distance  60  =  6  X  10  elevation  is 3 . 1440 

For  distance  500  =  5  X  100  elevation  is. .        26.2000 


Total 29.7108 

For  distance  567  ft.,  therefore,  elevation  is  29.71  ft. 


114 


THE  TRIGONOMETRIC  FUNCTIONS 


EXERCISES. 

1.  Check  the  above  method  of  interpolation  by  using  your  table  of 
logarithms  and  the  formula:  elevation  =  horizontal  distance  X  tangent 
of  the  angle. 

2.  Make  out  a  table  corresponding  to  the  table  above  replacing  hori- 
zontal distances  by  distances  measured  along  the  slope,  using  the  formula: 
elevation  =  distance  along  slope  X  sine  of  angle. 

3.  Compare  this  table  with  one  in  text.    What  conclusion  do    you 
draw  when  the  angle  is  small  ? 

In  solving  problems  connected  with  traverse  sailing  or  with 
land  surveying,  it  is  often  necessary  to  find  the  latitude  and 
departure  corresponding  to  the  several  courses  and  distances. 

Since  latitude     =  distance  X  cos  of  bearing, 

departure  =  distance  X  sin  of  bearing, 

it  is  convenient  to  use  a  table  in  which  these  projections  have 
been  computed.     Following  is  a  section  of  such  a  table: 

TRAVERSE  TABLE. 


Dist.  1. 

Dist.  2. 

Dist.  3. 

Dist.  4. 

Dist.  5. 

O           8f> 

On   r«w» 

Lat. 

Dep. 

Lat. 

Dep. 

Lat. 

Dep. 

Lat. 

Dep. 

Lat. 

Dep. 

30°  15' 

0.8638 

0.5038 

1.7277 

1.0075 

2.5915 

1.5113 

3.4553 

2.0151 

4.3192 

2.5189 

59°  45' 

30 

8616 

5075 

7233 

0151 

5849 

6226 

4465 

0302 

3081 

5377 

30 

45 

8594 

5113 

7188 

0226 

5782 

5339 

4376 

0452 

2970 

5565 

15 

31      0 

8572 

5150 

7142 

0301 

5716 

5451 

4287 

0602 

2858 

5752 

59     0 

15 

8549 

5188 

7098 

0375 

5647 

5563 

4196 

0751 

2746 

5939 

45 

30 

8526 

5225 

7053 

0450 

5579 

5675 

4106 

0900 

2632 

6125 

30 

45 

8504 

5262 

7007 

0524 

5511 

6786 

4014 

1049 

2518 

6311 

15 

32     0 

8480 

5299 

6961 

0598 

5441 

5S98 

3922 

1197 

2402 

6406 

58     0 

15 

8457 

5336 

6915 

0672 

5372 

6008 

3829 

1345 

2286 

6681 

45 

30 

8434 

5373 

6868 

0746 

5302 

6119 

3736 

1492 

2170 

6865 

30 

45 

0.8410 

0.5410 

1  6821 

1.0810 

2.5231 

1.6229 

3.3642 

2.1639 

4.2052 

2.7049 

15 

33     0 

8387 

5446 

6773 

0893 

5160 

6339 

3547 

1786 

1934 

7232 

57     0 

Dep. 

Lat. 

Dep. 

Lat. 

Dep. 

Lat. 

Dep. 

Lat. 

Dep. 

Lat. 

C/OUTS6. 

CoUTSG. 

Dist.  1. 

Dist.  2. 

Dist.  3. 

Dist.  4. 

Dist.  5. 

From  this  table  determine  the  latitude  and  departure  of  a 
Bourse  with  bearing  31°  10'  and  length  (distance)  413. 


SPECIAL  INTERPOLATION 


115 


Determine  the  latitude  and  departure  of  a  course  with  bear- 
ing 31°  10'  and  length  413. 

A  surveyor  runs  a  line  N.  30°  45'  E.,  distance  243,  then  N. 
32°  15'  W.,  distance  214.  How  far  north  and  how  far  east  or 
west  is  he  from  the  starting  point? 

By  use  of  the  traverse  table,  find  the  area  of  a  right  triangle 
which  has  a  hypotenuse  534  units  in  length  and  an  acute 
angle  59°  10'. 

A  table  of  logarithmic  and  natural  haversines  is  indispen- 
sable in  connection  with  problems  in  nautical  astronomy.  Be- 
low is  a  portion  of  such  a  table. 

LOGARITHMIC  AND  NATURAL  HAVERSINES 

hav  6  =  %  vers  6  =  $  (1  —  cos  6)  =  sin2  6/2. 


a. 

3"  50m  57°  30'. 

3h  51m  57°  45'. 

3h  52m  58°  0'. 

3»  53m  58°  15'. 

3h  54m  58°  30'. 

a. 

Log 
hav. 

Nat. 
hav. 

Log 
hav. 

Nat. 
hav. 

Log 
hav. 

Nat. 
hav. 

Log. 
hav. 

Nat. 
hav. 

Log. 
hav. 

Nat. 
hav. 

0 

9.36427 

.23135 

9.36772 

.23319 

9.37114 

.23504 

9.37455 

.23689 

9.37794 

.23875 

60 

1 

.36433 

.23138 

.36777 

.23322 

.37120 

.23507 

.37461 

.23692 

.37800 

.23878 

59 

2 

.36439 

.23141 

.36783 

.23325 

.37126 

.23510 

.37467 

.23695 

.37806 

.23881 

58 

3 

.36444 

.23144 

.36789 

.23329 

.37131 

.23513 

.37472 

.23699 

.37811 

.23884 

57 

+1' 

9.36450 

.23147 

9.36794 

.23332 

9.37137 

.23516 

9.37478 

.23702 

9.37817 

.23887 

56 

5 

.36456 

.23150 

.36800 

.23335 

.37143 

.23519 

.37484 

.23705 

.37823 

.23891 

55 

6 

.36462 

.23153 

.36806 

.23388 

.37148 

.23523 

.37489 

.23708 

.37828 

.23894 

54 

7 

.38467 

.23156 

.36812 

.23341 

.37154 

.23526 

.37495 

.23711 

.37834 

.23897 

53 

+2' 

9.36473 

.23160 

9.36817 

.23344 

9.37160 

.23529 

9.37501 

.23714 

9.37840 

.23900 

52 

20"  9™ 

20h  8m 

20"  7" 

20"  6m 

20"  5m 

Solve  the  following  problems  by  use  of  above  table  (do  not 
determine  6} . 

Log  hav  6  =  9.37144-10,  find  nat.  hav  6. 
Nat.  hav  6  =  0.23893,  find  log  hav  6. 

t  (time)  =  3h  51m  3s,  find  log  hav  t  and  nat.  hav  t. 

t  =  20h  7m  56s,  find  log  hav  t  and  nat.  hav  t. 
Find  log  hav  and  nat.  hav  of  58°  0'  25". 


116  THE  TRIGONOMETRIC  FUNCTIONS 


1.  The  angle  at  the  center  of  a  circle  is  4  radians.    The  arc  is  7'.    What 
is  the  radius  of  the  circle? 

2.  What  is  the  value,  hi  degrees  of  -=-  radians?    Of  -=-  radians? 

i  O 

3.  What  is  the  value  in  radians  of  40°?    55°?     1200°? 

4.  An  arc  is  5'.    The  radius  is  8'.    What  is  the  angle  at  the  center  in 
radians?    In  degrees? 

5.  One  angle  exceeds  another  by  -=-  radians.    The  sum  of  the  angles 

o 

is  175°.     Find  the  angles. 

6.  Prove  sin  a  cos  a  =  sin3  a  cos  a  +  cos3  a  sin  a. 

7.  Prove  cot  a  esc  a  =  I/ (sec  a  —  cos  o). 

8.  Prove  (1  +  tan2  a) /(I  +  cot2  a)  =  sin2  a/cos2  a. 

9.  Prove  sin  a  —  ski  o/(cot  a  —  1)  =  sin  a  [1  —  sin  a/ (cos  a  —  sin  a)]. 

10.  Given  tan  30°  =  $  Vjj,  find  tan  15°,  by  use  of  a  formula. 

11.  Using  the  addition  theorem  find  sin  (a  +  b  +  c). 

12.  Prove  cos  2  a  =  (1  -  tan2  o)/(l  +  tan2  o). 

13.  Prove  sin  f  j  —  a  1  =  (cos  a  —  sin  a)/ V2. 

14.  Prove  tan  a  +  tan  b  =  sin  (o  +  6)  /cos  a  cos  &. 

15.  Prove  cos  a  =  (1  -  tan2  i  o)/(l  +  tan2 1  a). 

16.  Prove  cos  (45°  +  a) /cos  (45°  -  a)  =  sec  2  a  -  tan  2  a. 
Solve  and  determine  all  values  less  than  360°. 

17.  tan  a  =  sin  a. 

18.  (3-4  cos2  a)  cos  2  a  =  0. 

19.  sin  a  +  cos  a  cot  a  =  2. 

20.  tan  (45°  +  o)  =  3  tan  (45°  -  o). 

21.  5  ski  a  =  tan  a. 

22.  2  cos  a  +  sec  a  =  3. 

23.  2  sin  a  +  5  cos  a  =  2. 

24.  tan  2  a  =  1. 

25.  If  tan  2  o  =  m,  find  tan  o. 

26.  If  cos  o  =  J,  find  cos  J  a. 

1  ^ 

27.  Prove  arc  tan  ^  +arc  tan  c  =  T>  given  tan  -.  =  1. 

28.  Solve  for  y  in  2  arc  sin  |  +  arc  sin  y  =  w/2. 

29.  In  the  figure  the  following  data  are  given  to  find  AB. 

CD  =  943.4;  CE  =  673.3;  a  =  72°  9.3';  ft  =  60°  17.9';  y  =  32°  14.6'; 
5  =  67°  33.9';  e  =  19°  14.7'.    See  Fig.  54. 

Ans.  AB  =  1054. 

30.  A  coal  mine  entry  runs  N .  1 8°  W .  to  A  whose  coordinates  are  (450, 100) . 
A  post  on  the  boundary  line  has  coordinates  (575,  475) .    The  bearing  of  the 


MISCELLANEOUS  EXERCISES  117 

boundary  line  is  N.  26°  50'  E.  Another  entry  has  been  run  N.  80°  E.  to  a  point 
B  across  the  boundary  whose  coordinates  are  (112,  174).  It  is  desired  to 
continue  the  first  entry  to  a  point  10'  from  the  boundary,  then  parallel  to  the 
boundary  to  a  point  where  the  second  entry  would  intersect  it  if  continued, 


FIG.  54. 

then  to  follow  a  line  that  would  coincide  with  the  "continuation  of  the 
second  entry.  Make  the  calculations  and  determine  thejcoordinates  of  the 
points  of  turning. 

31.  Find  the  coordinates  of  P  from  the  following:  Start  at  A  whose  co- 
ordinates are  (-75,  350);  run  S.  26°  15'  W.,  355';  then  run  S.  54°  20'  E., 
175';  then  run  N.  10°  15' E.,  300';  then  run  N.  75°  45'  W.,  500'  to  P. 

32.  Since  the  mil  is  the  angle  at  the  center  of  a  circle  subtended  by 
•sdrs  of  the  circumference,  therefore, 

Measure  in  mils  of  angle  A     _  6400  _  160 
Measure  in  degrees  of  angle  A       360         9 

Change  the  following  angles  in  degrees  to  mils:  45°,  90°,  135°,  180°,  225°, 
270°,  315°,  360°. 

By  means  of  the  definition  of  a  mil  and  that  of  a  radian  determine  the 
conversion  factor  for  changing  angles  from  mils  to  radians.  Change  the 
following  angles  in  radians  to  mils:  ir/2,  ir/4,  if. 

Change  the  following  angles  in  mils  to  radians:  800,  1600,  2400,  3200. 

33.  By  the  definition  of  a  mil,  if  the  radius  of  a  circle  is  1000  yds.,  an 
arc  of  1  yd.  subtends  an  angle  at  the  center  of  approximately  one  mil.    In 


118 


THE  TRIGONOMETRIC  FUNCTIONS 


the  figure,  therefore,  from  similar  triangles,  assuming  /  and  arc  s  equal, 
s       1000 

rki*        o   =    — 

R 


1000    „ 
or    s  =  -5-  •  F, 


F        R 

or,  since  the  angle  t  in  mils  has  the  same  measure  as  the  arc  s, 

1000 


t  = 


H 


F. 


Note.  —  The  assumptions  made  in  the  above  discussion  give  results 
sufficiently  accurate,  in  general,  for  work  in  gunnery  where  t  is  usually  a 
very  small  angle. 

(o)  If  the  range  R  =  1000  yds.,  and  the  distance  between  the  two  guns 
F  =  20  yds.,  what  is  t  in  mils? 

G, 


-1000- 


FIG.  55. 
What  is  t  if 
(6)  R  =  1000yds.,  and  F  =  40yds.? 

(c)  R  =  1000  yds.,  and  F  =  n  yds.  ? 

(d)  R  =  2000  yds.,  and  F  =  20  yds.  ? 

(e)  R  =  2500  yds.,  and  F  =  30yds.? 

34.   The  angle  at  the  target  T  (or  aiming  point  P)  subtended  by  the 

distance  between  two  successive 
pieces  G  and  G\  on  the  battery  front 
is  called  the  parallax  of  the  target  (or 
aiming  point).  If  the  aiming  point 
is  in  the  line  of  the  battery  front  the 
parallax  of  the  point  is  a  minimum, 
zero.  If  the  target  (aiming  point)  is 
on  the  normal  to  the  battery  front, 
GGi  at  the  mid-point  the  parallax  is  a 
maximum.  This  is  called  the  normal 
parallax.  In  practice,  if  the  target  is 
on  any  normal  to  the  battery  front, 
In  what  follows  the  normal  through 


FIG.  56. 


its  parallax  is  considered  as  normal. 
G  is  chosen. 


MISCELLANEOUS  EXERCISES 


119 


The  normal  parallax  t  of  T  is  (see  exercise  33,  above) 
GGl  F 


/e/iooo 


where  R  is  the  distance  from  T  to  battery  front. 

(a)  What  is  the  normal  parallax  of  the  target  for  a  range  2500  yds.,  and 
distance  20  yds.  between  G  and  GI? 

(6)   What  is  the  normal  parallax  if  R  =  500  yds.  and  F  -  40  yds.? 

35.  If  the  target  is  not  on  the  normal  to  the  battery  front,  a  correction 
for  "obliquity"  is  required.  This  is  called  "correction  of  parallax  due  to 
obliquity." 

The  angle  at  G  between  the  line  normal  to  the  battery  front  and  the  line 
to  the  target  T'  is  the  angle  of  obliquity.  In  the  figure,  0  is  the  angle  of 
obliquity  of  the  target  T'. 

The   true   parallax   (t)   is  the  normal  T 

parallax  after  it  is  corrected  for  obliquity.  A  """^"Vr' 

True  parallax  is  obtained  as  follows: 

DG  =  GiG  cos  DGGi,  approximately, 


But 

t'  (mils)  =  DG  ^ .     (See  Exercise  33.) 
Therefore 

t'  (mils)  = 


=  t  (mils)  cos  0, 


1000  **1  G 

GiG  — =-  =  normal  parallax  of  T.  FIG.  57. 

Therefore,  true  parallax  =  normal  parallax  X  cosine  of  the  angle  of 
obliquity. 

(o)  If  the  normal  parallax  t  =  8  mils,  and  the  obliquity  angle  is  600 
mils,  find  the  true  parallax. 

(6)  If  R  =  2500  yds.,  the  obliquity  angle  600  mils,  and  G,G  =  20  yds., 
find  the  true  parallax. 

(c)   Verify  the  formula,  true  parallax  t'  =   r  '  ^L  ,  where  angle  0  is  as 

K/  \\j\j\j 

indicated  in  the  figure. 


CHAPTER  IX 

POLAR  COORDINATES,   COMPLEX  NUMBERS, 
VECTORS 

73.  The  position  of  a  point  can  be  determined  by  means  of 
a  distance  and  an  angle.  Let  OX  be  a  fixed  reference  line,  called 
the  axis.  Call  0  the  pole.  Then  if  the  angle  6,  and  the  distance 
OP  =  r,  are  known  the  position  of  P  is  known.  The  point  P 
is  designated  as  P  (r,  6)  or  as  (r,  0).  The  number  pair  (r,  0)  are 
the  polar  coordinates  of  P.  OP  =  r  is  the  radius  vector  of  P, 
and  0  is  the  vectorial  angle  of  P. 


FIG.  58. 

If  P  lies  on  the  terminal  line  of  the  vectorial  angle,  r  is  reckoned 
as  positive.  If  P  lies  on  the  terminal  line  of  the  vectorial  angle 
produced  on  the  opposite  side  of  the  pole,  r  is  regarded  as  nega- 
tive. The  vectorial  angle  6  is  positive  or  negative  according 
as  it  is  reckoned  counter  clockwise  or  clockwise,  respectively. 
In  the  figure,  if  P'  is  regarded  as  P'  (r'd"),  r'  and  0"  are  positive. 
If  P'  is  regarded  as  (r',  6'),  r'  is  negative.  If  P'  is  regarded  as 

120 


POSITION 


121 


(r'0'"),  r'  is  positive  and  6'"  is  negative,  and  similarly  in  other 
cases. 


1.   Locate  the  following  points  (4,  jj,    (6,30°),    [—5,  o)> 


2.  Construct  graph  of  r  =  8  6 
(6  in  radians). 

Note.  —  Arrange  a  table  of 
values  as  in  drawing  graphs  on 
rectangular  coordinates.  Then 
lay  off  the  points  as  above  and 
draw  a  smooth  curve  through 
the  points. 


Assume 


=  0, 


TT 

6' 


0 
FIG.  59. 


7T  7T 

3'       2' 


Calculate 


rv 

=  0, 


o 

g,  8-3, 


A  portion  of  the  graph  is  shown  in  the  figure. 

3.  Construct  the  graph  of  r  =  8  cos  0. 

4.  "  " 

5.  "  * 

6.  "  * 

n  (i                            <i 

8.  "  " 

9.  "  " 


10. 


r  =  2/(l  —  cos0). 

r  =  sin  2  6. 

r  sin  6  =  4  (solve  for  r). 

r  =  4/(l  -3cos0). 

r  =  7  (a  circle). 

r  =  cos  3  0. 

0 
r  =  cos  -• 


11.  In  2  x  +  3  y  =  5,  put  x  =  rcosd,y  =  rsind  and  draw  the 
graph  of  the  resulting  equation.     Draw  the  graph  from  the 
original  equation. 

12.  In  y2  =  8  x,  put  y  =  r  sin  0,  x  =  r  cos  0  +  2.     Note  the 
form  of  the  resulting  equation.     Construct  the  graphs  of  both, 
equations. 


122      POLAR  COORDINATES,  COMPLEX  NUMBERS,  VECTORS 

74.  In  the  solution  of  certain  quadratic  equations  and 
equations  of  higher  degree,  there  occur  roots  of  the  form  a  +  bi, 
where  a,  b  are  real  and  i2  =  —  I  or  i  =  V— 1,  36gr. 

From  the  above  definition  of  i  it  is  easily  found  by  calculating 
that, 

i  —  v     i, 
^  =  (V^I)2  =  -1, 
#  =  #.{  =  (_i)  V^T  =  -V^i  =  -i, 


The  value  of  i5  is  the  value  of  i,  the  value  of  i6  will  be  the  value 
of  i2,  etc. 

75.  A  geometric  basis  for  interpreting  complex  numbers  is 
attributed  to  Argand.  Since  any  number  or  any  line  segment 
has  its  sign  or  direction  reversed  when  multiplied  by  —  1  =  i2, 
the  line  is  rotated  180°  counter  clockwise  by  multiplying  by 
-1  =  i2.  Thus  in  the  figure  AB  -  (-1)  =  AB  •  V  =  -AB  = 
AB'.  It  then  looked  reasonable  to  suppose  that  AB  is  rotated 
90°  counter  clockwise  when  multiplied  by  V— 1  =  i  or  AB  is 

just  half  reversed.  Thus  AB  • 
i  =  AB"  in  the  figure.  This 
idea  proves  to  be  very  useful. 
The  number  a  +  bi  may  now 
be  looked  upon  as  a  point  in 
the  plane  determined  by  two 
steps  taken  perpendicular  to 
each  other.  Thus,  if  a  =  Om 
and  b  =  mn  in  Fig.  61,  a  +  bi 
will  be  Om  +  mP,  since  bi  is 
perpendicular  to  6.  This  idea 
is  then  equivalent  to  regarding  (a,  6)  in  a  +  bi  as  the  rectan- 
gular coordinates  of  a  point  P. 

Polar  coordinates  may  also  be  associated  with  P.     Thus,  if 
angle  XOP  =  6,  and  OP  =  r,  P  may  be  regarded  as  P  (r,  6). 
The  line  OP  =  r  is  called  the  modulus  or  absolute  value  of 


FIG.  60. 


ARITHMETIC  OPERATIONS  WITH  COMPLEX  NUMBERS      123 


the  number  a  +  bi.    The  angle  6  is  called  its  amplitude, 
the  figure  it  is  easy  to  see  that  for  any  position  of  P, 


From 


(1) 
(2) 

(3) 
(4) 


r  =  Va?  +  b2. 


sin0  =  -• 
r 

COS0  =  -« 

r 


tan  6  =  — 
a 


FIG.  61. 


These   relations   are    funda- 
mental. 

1.  Locate  on  rectangular 
coordinate  paper  the  points, 

2  + 3  i  (o  =  2,  b  =  3);    2- 
3i;    1— 4i;    — 4— 3i;    —5 
+  2i. 

2.  Find  the  modulus  and  amplitude  of  each  number  in  (1). 
76.  Arithmetic  operations  with  complex  numbers.  — 

(a)  Two  complex  numbers  are  equal  when  and  only  when 
they  represent  the  same  point  referred  to  the  same  axes.     Hence 
the  two  complex  numbers  a  +  bi  and  a'  +  b'i  are  equal  when 
and  only  when  a  —  a'  and  b  =  b'. 

(b)  The  sum  of  two  complex  numbers  a  +  bi  and  a'  +  b'i  is 
the  complex  number  a  +  a'  +  (6  +  b')  i.     Thus  the  sum  of 

3  +  2i  and  1  +  4i  is  4  +  Qi. 

1.  Add  3  +  5  i  to  1  +  2  i;   I  +  6  i  to  3  -  2  i;    -£  -$  i  to 
— ?  +  $  i,  locate  each  point  and  each  sum  on  a  diagram  in  rec- 
tangular coordinates. 

2.  Add  ( -2  +  3  i),  ( - 1  -  2  i),  (3  +  6  i).    Locate  all  points 
and  the  sum. 

3.  Add  (1  +  3  i),  (-3  -  i),  (6  +  7  i).    Locate  all  points  and 
the  sum. 

(c)  Multiplicatibn  of  two  complex  numbers  is  defined  by 

(a  +  bi)  (a'  +  b'i')  =  aa'  -  bb'  +  (ab'  +  a'b)  i. 
Note.  —  Remember  in  carrying  out  the  work,  i2  =  —  1. 


124      POLAR  COORDINATES,  COMPLEX  NUMBERS,  VECTORS 

(d)   Division  of  complex  numbers  is  defined  by 
a  +  bi        (a  +  bi)  (a'  -  b'i)  =aa'  +  W      (a'b  -  ob')  i 
a'  +  b'i      (a1  +  b'i)  (a'  -  b'i)       a'*  +  b'2  H      a'2  +  b'2 

The  operations  (6),  (c),  (d)  are  in  general  possible  and  lead  to 
complex  numbers,  impossible  if  a'2  +  b'2  =  0,  indeterminate  if 
a*  +  b'2  =  a2  +  b2  =  0,  see  42,  43,  77. 

1.  (3-4i)(7  +  4i)  =?     (3-4i)(-7-4i)  =  ?    Locate 
all  points  and  the  products. 

2.  (3  +  4i)-5-(7  +  4i)  =  ?  (3-4i)-r-(-7-4i)  =  ?    Locate 
all  points  and  the  quotients. 

3.  (2  +  3i)-5-(l-r)  =  ?   (l-i)   (l+i)-5-2  +  2i  =  ?    Locate 
all  points  and  the  quotients. 

77.   By  the  formulas  of  73,  a  +  bi  may  be  put  in  the  form 

a  +  bi  =  Vtf+W-=J=  +     .U 

\Va2  +  b2      Va2  +  b2 

=  r  (cos0  +  isin0). 

The  last  form  is  called  the  polar  form  of  the  complex  number 
(a  +  bi),  where 

r  =  Va2  +  b2,  sin  6  =     .  and  cos  0  = 


f CVJ.J.V4.         \J\JYJ    I/       —  / 

'a2  +  b2  Va2  +  b2 

Multiplication  and  division  of  complex  numbers  take  very 
interesting  and  useful  forms  in  polar  coordinates.     Thus 
(a  +  bi)  (a'  +  b'i)  =  r  (cos  d  +  i  sin  6)  r'  (cos  tf  +  i  sin  00 

=  rr'  [(cos  6  cos  0'  -  sin  0  sin  0')  +  i  (sin  0  cos  0'  +  cos  0  sin  0')] 
=  rr'  [cos  (0  +  0')  +  i  sin  (0  +  0')]. 

This  shows  that  the  modulus  of  the  product  is  the  product  of 
the  moduli  and  the  amplitude  of  the  product  is  the  sum  of  the 
amplitudes  of  the  numbers  multiplied. 
Again, 

a  +  bi'       r  (cos  0  +  i  sin  0)  (cos  6'  —  i  sin  0') 
a'  +  b'i  ~  r'  (cos  8'  +  i  sin  0')  (cos  0'  -  i  sin  0') 

=  -,  [(cos  (0  -  0')  +  i  sin  (0  *-  0')]. 

T 

This  shows  that  the  modulus  of  the  quotient  is  the  modulus  of 


VECTORS 


125 


the  dividend  divided  by  the  modulus  of  the  divisor  and  the 
amplitude  of  the  quotient  is  the  amplitude  of  the  dividend 
minus  the  amplitude  of  the  divisor. 

As  an  exercise  reduce  each  complex  number  in  the  last 
section  to  polar  form  and  perform  the  indicated  operations  by 
use  of  the  above  formulas. 

78.  There  are  a  number  of  physical  quantities,  of  fundamental 
importance,  such  as  force,  velocity,  electric  field,  etc.,  that  are 
completely  specified  by  magnitude  and  direction  of  the  line  of 
action.     They  are  called  vector  quantities  in  distinction  from 
non-directed  quantities,  such  as  energy,  speed,  etc.     The  latter 
are  called  scalar  quantities  for  the  reason  that  they  are  com- 
pletely specified  by  magnitude  alone. 

A  directed  straight  line  segment  is  a  vector.  It  is  evident 
that  a  vector  may  represent  a  vector  quantity.  For  the  length 
of  the  vector,  to  some  Y 

scale,  may  represent  the 
magnitude  of  the  vector 
quantity  and  the  vector 
may  be  parallel  to  the  line 
of  action  of  the  vector 
quantity. 

If  A  B  is  a  vector,  its 
direction  is  understood  to 

be  from  A  toward  B.    A  

is  the  initial  point  of  the 
vector  AB,  and  B  is  the 
terminal  point. 

A  vector  may  be  dis- 
placed parallel  to  itself 
without  affecting  its  mag- 
nitude or  direction.  When 
a  vec-tor  may  be  so  moved  it  is  called  a  free  vector.  When  a 
vector  is  attached  to  a  fixed  point,  it  is  a  fixed  or  localized  vector. 

79.  The  notation  of  complex  numbers  and  of  polar  coordi- 
nates lends  itself  readily  to  representing  vectors  and  to  cal- 


FIG.  62. 


126      POLAR  COORDINATES,  COMPLEX  NUMBERS,  VECTORS 

culating  with  them.  Thus  the  vector  OP  in  Fig.  62  is  de- 
scribed either  as  (x  +  iy)  or  as  (r,  6)  at  pleasure.  Since  a 
vector  symbolizes  a  vector  quantity,  any  operations  of  arith- 
metic performed  with  vectors  will  have  a  similar  meaning  with 
vector  quantities.  The  direction  of  OP  is  that  from  0  to  P. 
When  OX  is  the  reference  line  (axis)  the  angle  6  determines 
the  direction  of  OP. 

80.  Addition  and  subtraction  of  vectors.  —  The  most  con- 
venient notion  from  which  to  derive  vector  addition  and 
subtraction  is  displacement  or  step.  Thus  to  add  a  vector 
BC  =  b*  to  another  vector  AB  =  a,  lay  off  a  and  then  from  the 


FIG.  63. 


terminal  point  of  a  lay  off  b.  The  vector  from  the  initial  point 
of  a  to  the  terminal  point  of  b,  that  is,  AC  =  c,  is  the  vector  sum 
of  a  and  b.  This  may  be  expressed  by  the  vector  equation 

a  +  b  =  c. 

That  is,  a  displacement  AB  —  a  followed  by  the  displacement 
BC  =  b  is  equivalent  to  the  displacement  AC  =  c. 

If  the  vector  B'C'  is  to  be  subtracted  from  the  vector  AB', 
lay  off  the  vector  B'C"  equal  in  length  to  B'C'  but  in  the  oppo- 
site direction.  The  vector  AC"  is  the  difference.  If  AB'  = 
a'j  B'C'  =  b'  and  AC"  =  c'  we  may  write  the  vector  equation 

a'  -  b'  =  c'. 
*  When  a  single  letter  represents  a  vector  heavy  type  is  used. 


ADDITION  AND  SUBTRACTION  OF  VECTORS         127 


It  is  seen  that  to  subtract  a  vector  is  the  same  as  to  add  its 
opposite  or  negative. 

To  add  several  vectors  OP, 
OP',  OP",  ...  lay  off  the 
vectors  successively,  with  in- 
itial point  of  each  on  the 
terminal  point  of  the  preced- 
ing, forming  the  sides  of  a 
polygon.  The  closing  side 
of  the  polygon  drawn  from 
the  starting  point  to  the 
terminal  point  of  the  last 
vector  is  the  vector  sum  of  the  given  vectors,  Fig.  65.  The 
vector  equation  is 

OP"  +  OP  +  OP'  =  O'P'. 


FIG.  64. 


FIG.  65. 

It  must  be  remembered  we  are  not  adding  the  lengths  of  the 
lines  alone  but  the  displacements  (including  directions  of  the 
lines).  The  sum  of  two  sides  of  a  triangle  is  vectorially  equal 
to  the  third  side,  but  not  numerically. 

1.  Find  the  vector  sum  of  two  adjacent  sides  of  a  parallelo- 
gram ABCD.    Find  the  vector  difference  of  the  same  sides. 

2.  Find  the  vector  sum  of  the  three  sides  of  a  triangle  ABC, 


128      POLAR  COORDINATES,  COMPLEX  NUMBERS,  VECTORS 

taken  in  order.    Find  the  vector  difference  AB  —  BC.    Find 
the  vector  sum  AB  +  BC. 

3.  Find  the  sum  of  the  vectors  whose  initial  points  are  at  the 
pole  and  whose  terminal  points  are  (3,  60°)  ;  (10,  —20°)  ;  (—24, 
120°),  respectively. 

4.  Find  the  sum  of  the  vectors  represented  by  (3  —  6  i)  ; 
(4  +  2t);  (5-3t);  (l-4t). 

5.  Find  the  magnitude  and  direction  of  the  velocity  resulting 
from  two  simultaneous  velocities,  one  due  north  50'/sec.,  the 
other  N.  45°  E.,30'/sec. 

Note.  —  By  75,  76  these  velocities  may  be  expressed  in  rec- 
tangular form  and  added.  The  sum  can  then  be  reduced  to 
polar  coordinates.  The  result  may  be  found  by  use  of  the  law 
of  58.  Solve  by  both  methods  and  check  the  results. 

6.  A  boat  is  rowed  across  the  current  of  a  river  at  5  mi./hr. 
The  current  is  1|  mi./hr.     Find  the  actual  velocity  of  the  boat 
in  magnitude  and  direction. 

7.  If  a  horse  is  running  N.  35°  E.  at  the  rate  of  12  mi./hr., 
how  fast  is  he  going  north  ?     How  fast  east  ? 

8.  Make  diagrams  to  scale  for  Exs.  6,  7  and  verify  your 
calculations. 

9.  Can  you  explain  by  the  vector  idea  why  a  division  of 
society  into  opposing  factions  retards  or  prevents  social  progress? 

10.  Show  by  vector  addition  the  truth  of  the  parallelogram 
of  forces. 

81.*  Product  of  Two  Vectors.  —  Vectors  exhibit  two  types 
of  product: 

(a)  Scalar  product  of  two  vectors  is  defined  as  the  product 
of  their  magnitudes  and  the  cosine  of  their  included  angle. 
Thus  if 

F  =  r  (cos  e  +  i  sin  0)  (77,  79) 
and 

S  = 


*  The  remainder  of  this  chapter  may  be  omitted  if  desired.  It  is 
recommended,  however,  if  the  time  permits  and  the  student's  knowledge 
of  mechanical  notions  justifies,  that  the  entire  chapter  be  covered  carefully. 


PRODUCT  OF  TWO  VECTORS          129 

the  scalar  product  of  F  and  S  will  be  written  as 

FS  =  rr'  cos  (0'  -  0)  =  FS  cos  (0'  -  0). 

Since  cos  (— a)  =  cos  a,  the  order  of  taking  the  factors  is  in- 
different. 

(6)  Vector  product  of  two  vectors  is  defined  as  the  product 
of  their  magnitudes  and  the  sine  of  their  included  angle.  The 
vector  product  of  F  and  S  will  be  written  as  vFS  and  is  numeri- 
cally equal  to 

FS  sin  (0'  -  0)*  =  rr'  sin  (0'  -  0), 

where  7^,  S  are  the  magnitudes  of  F,  S,  respectively.     As  the 
name  indicates  the  vector  product  is  a  vector.     The  direction 
is  that  of  the  travel  of  a  right-hand  screw, 
perpendicular  to  the   plane  of  the  two 
vectors,  when  turned  in  such  a  way  as 
to  rotate   the   first   factor   toward   the 
second.     Then  0'  —  0  is  regarded  posi- 
tive.    If  the  direction  of  rotation  is  re- 
versed the  angle  0'  —  0  becomes  negative 
and  its  sine  becomes  negative  and  con- 
sequently the  vector  product  changes  sign 
and  its  vector  is  reversed.     The  order  of 
factors  in  the  vector  product  is,  therefore,  not  indifferent,  but 
must  be  carefully  noted. 

1.  Find  the  scalar  product  of  (3,  40°)  and  (6,  30°).     Find  the 
vector  product,  factors  taken  in  order  given. 

Note:  30°  -40°  =  -10°. 

2.  Find  the  scalar  and  vector  products  of  (25,  35°)  and  (60, 
124°)  in  order  given. 

3.  If  work  is  defined  as  the  product  of  force  multiplied  by 
the  displacement  component  in  the  direction  of  the  force,  show 
that  work  is  the  scalar  product  of  force  and  displacement. 

*  It  is  noted  that  8  sin  (&'  —  0)  is  the  perpendicular  distance  from  the 
origin  to  a  line  through  the  terminal  of  S  parallel  to  F*  This  distance,  in 
case  .F*  is  a  force,  is  called  the  moment  arm  of  F  about  the  origin  as  an  axis. 
See  Ex.  5  below. 


130      POLAR  COORDINATES,  COMPLEX  NUMBERS,  VECTORS 


4.  What  is  the  work  done  by  a  force  of  2000  Ibs.  acting 
N.  30°  E.  in  moving  a  car  50'  on  a  track  due  north?     (No 
friction.) 

5.  The  moment  of  a  force  about  a  point  or  axis  is  defined  to 
be  the  product  of  the  force  by  the  perpendicular  distance  of  the 
line  of  the  force  from  the  point  or  axis.     Show  that  the  moment 
of  a  force  about  any  axis  is  the  vector  product  of  the  vector 
distance  from  the  line  of  the  force  to  the  axis  and  the  vector 
force.     Note  the  order  of  factors. 

6.  Find  the  moment  of  a  force  of  500  Ibs.  acting  N.  40°  E. 
about  a  point  such  that  the  vector  from  the  point  to  the  initial 
point  of  the  force  vector  is  12  units  in  a  direction  N.  80°  E. 

7.  The  coordinates  of  the  initial  point  of  a  force  vector  are 
(3,  6),  the  terminal  point  (4,  10).     Find  the  moment  of  the 
force  about  the  origin.     The  magnitude  of  the  force  is  the 
length  of  the  line  joining  the  above  two  points.     The  moment 
arm  is  the  distance  of  the  line  from  the  origin. 

8.  Find  the  scalar  product  of  (16,  30°)  by  (24,  60°). 

9.  Find  the  vector  product  of  the  vectors  in  Ex.  8. 

Note.  —  We  shall,  from  now  on,  describe  a  vector  having  its 
initial  point  at  the  origin  by  giving  the  coordinates  of  its  ter- 
minal point  only. 

82.  A  vector  may  be  re- 
garded as  the  vector  sum 
of  its  components  on  the 
axes.  Thus  the  vector  OP 
is  the  vector  sum  of  Om  and 
mP.  We  shall  adopt  the 
notation  of  writing  a  sub- 
script to  indicate  the  com- 
ponent. Thus  the  vector 
OP  will  be  denoted  by  P 
and  its  component  on  OX 
and  its  component  on  OF  by  Py.  The  vector  equation 


o 


FIG.  68. 


holds  for  all  vectors. 


P  = 


VECTOR  PRODUCT 


131 


The  scalar  and  vector  products  of  two  vectors  can  now  be 
expressed  in  rectangular  coordinates.  If  F  =  F^  +  FV}  S  = 
So,  +  Sv  we  may  write: 

(1)  FS  =  FA  +  FVSV. 

Now  if  F  is  a  force  and  S  a  displacement  then  F«,  is  a  force  and 
Sg,  a  displacement  in  the  same  direction  and  FJS,,,  is  the  work 
done  by  FO,  in  the  displacement 
So,,.  Similarly  FVSV  is  the  work 
of  Fy  in  the  displacement  Sy. 
The  right  side  of  (1)  then  is  the 
measure  of  work  done  by  the 
components  of  F.  But  work  is 
the  product  of  force  and  the 
component  of  displacement  in 
the  direction  of  the  force, ''that 
is,  the  product  of  force  and  dis- 
placement and  the  cosine  of 
their  included  angle.  There- 
fore, the  scalar  product  on  the 
left  of  (1)  represents  the  same  thing  that  the  right  side  repre- 
sents and  the  two  members  of  (1)  are  but  different  ways  of 
expressing  the  scalar  product  of  F  and  S. 

(2)  vSF  =  SXFV  -  SVF^  a  vector. 

The  term  SmFv  is  the  moment  of  the  force  Fy  acting  at  A  and 
tending  to  swing  A  counterclockwise  about  0.  The  term 
SvFa.  is  the  moment  of  Fx  acting  at  A  tending  to  swing  A  about 
0  clockwise  and  is  negative.  Strictly  speaking  the  right  side  of 
(2)  is  a  scalar  quantity,  but  owing  to  the  known  conventions 
as  to  the  direction  of  rotation  it  contains  the  means  of  determin- 
ing the  direction  of  the  tendency  to  rotation  and  we  are  justified 
in  calling  it  a  vector.  The  left  side  of  (2)  is  known  to  be  the 
moment  of  F  about  0  when  applied  at  A,  81.  Hence  the  two 
sides  of  (2)  represent  the  same  thing.  We  infer  the  generality 
of  (1),  (2),  for  all  vectors.  These  products  play  an  important 
role  in  physics  and  mechanics. 


FIG.  69. 


132      POLAR  COORDINATES,  COMPLEX  NUMBERS,  VECTORS 

1.  Find  the  axial  components  of  (35,  26°);    (125,   -65°); 
(-275,  -120°). 

2.  Find  the  sum  of  the  components  along  the  z-axis  and  the 
sum  of  the  components  along  the  y-axis  in  (1).     Consider  these 
sums  as  components  of  a  new  vector  and  calculate  its  modulus 
and  amplitude  (75). 

3.  Find  the  modulus  and  amplitude  of  the  vector  whose 
components  are  Pa,  =  —496,  Pv  =  275. 

4.  Find  the  moment  of  P«,  =  50,  Pv  =  75  applied  at  the 
point  (10,  8). 

5.  If  F  =  30  +  i  50  and  S  =  8  +  7  i,  find  the  scalar  and 
vector  products  of  F  and  S. 

6.  Reduce  (25,  120°)  and  (80,  40°)  to  rectangular  coordinates 
and  find  the  scalar  product  and  the  vector  product. 

7.  In  (6)  find  the  scalar  and  vector  products  without  reduc- 
ing to  rectangular  form. 

8.  Find  the  total  moment  about  the  origin  of  the  vectors 
(6  +  10  i),  (-4  +  3  i),  applied  at  the  point  (-3,  7). 

9.   A  lever  has  weights 

i  An 

as  shown  in  the  diagram. 
Find  the  distance  from  A 
to  the  point  of  application 
of  a  100  Ib.  force  that 
will  just  balance  the  other 
forces. 

Note.  —  The  sum  of  the 
v      -n  moments  of  the  downward 

£  IG.    /  U. 

forces  about  A  must  equal 

the  moment  of  100  Ibs.  about  A  where  x  is  the  unknown 
moment  arm.     That  is, 

lOOz  =  5  •  50  +  11  •  100  +  19  •  25, 

to  find  x. 

83.  The  notation  of  vectors  will  now  be  applied  to  some 
problems  in  the  equilibrium  of  particles  and  of  rigid  bodies 
acted  upon  by  external  forces. 


r- 


"T 


EQUILIBRIUM   OF  PARTICLES  133 

A  particle  is  a  geometric  point  regarded  as  having  inertia  or 
mass. 

A  rigid  body  is  one  of  finite  fixed  magnitude  and  unvarying 
form. 

(a)  Equilibrium  of  a  particle.  —  In  order  that  a  particle  shall 
not  have  its  state  of  rest  or  of  uniform  motion  in  a  straight  line 
changed,  all  forces  acting  on  the  particle  must  be  balanced. 
That  is,  there  must  be  no  component  of  resultant  force  along 
any  line  or  axes  of  reference.  Since  the  forces  may  be  repre- 
sented by  vectors,  this  means  the  vector  sum  of  all  forces  acting 
on  the  particle  must  be  zero.  This  in  turn  means  the  polygon 
formed  by  the  vectors  in  succession  must  be  a  closed  polygon. 
This  polygon  is  called  the  vector  polygon  of  forces. 

If  the  vector  sum  of  the  forces  acting  on  a  body  is  not  zero 
there  will  be  a  resultant  force  equal  to  the  closing  side  of  the 
polygon  which  is  the  vector  sum  of  the  vectors  of  the  forces. 
The  calculation  of  the  resultant  force  or  vector  sum  of  a  set  of 
forces  is  an  important  problem. 

(6)  A  rigid  body  is  in  equilibrium  when  it  is  at  rest;  moving 
uniformly  in  a  straight  line,  without  rotation;  rotating  uni- 
formly about  a  fixed  axis  or  moving  uniformly  in  a  straight  line 
and  rotating  about  an  axis  whose  direction  is  fixed. 

For  equilibrium  of  a  rigid  body  (a)  the  conditions  of  equilib- 
rium of  a  particle  must  be  satisfied;  (6)  the  sum  of  all  moments 
acting  on  the  body  must  be  zero,  when  taken  about  any  axis. 

The  solution  of  a  problem  relating  to  a  rigid  body  is  generally 
in  two  parts,  viz. :  First,  consideration  of  the  forces  as  acting  on 
a  particle  of  the  same  mass  as  the  body;  second,  determination 
of  the  moments  and  the  resultant  moment  acting  on  the  body. 

Some  examples  will  illustrate  how  problems  in  equilibrium  of 
particles  and  rigid  bodies  may  be  solved  in  ordinary  cases. 

I.  Let  a  particle  m  be  acted  on  by  the  forces  P,  Qt  R,  all  in 
the  same  plane  as  shown  in  Fig.  71.  To  find  the  resultant 
force.  P  =  75,  Q  =  100,  R  =  125. 

The  vector  equation  for  the  resultant  is: 

Resultant  =  S  =  R  +      +  P. 


134      POLAR  COORDINATES,  COMPLEX  NUMBERS,  VECTORS 


To  calculate  the  magnitude  and  direction  of  the  resultant, 
resolve  along  two  perpendicular  axes,  mX  and  mY.     Along  mX, 

Sx  =  S  cos  6  =  R  cos  120°  +  Q  cos  10°  +  P  cos  ( -20°) 
or  =  125  •  (-0.5)  +  100  •  (0.985)  +  75  •  (0.939) 

=  -62.5  +  98.5+70.4. 
/.     Sx  =  106.4. 

T 


FIG.  71. 
Along  mY, 

Sv  =  S  sin  6  =  72  sin  120°  +  Q  sin  10°  +  P  sin  (-20°) 
=  125  •  (0.866)  +  100  •  (0.173)  +  75  •  (-0.342) 
=  108  +  17.3  -  25.6. 
.'.     S,  =  99.7. 

=  V&2  +SV2  =  V106.42  +  99.72  =  141.3 
S»      99.7 


Now 
and 


=  0.938. 


.-.     6  =  43°  10'  (see  figure,  ms  =  S). 

In  a  similar  manner  an  unknown  force  in  any  system  acting 
on  a  particle  can  be  determined  if  the  equations  of  equilibrium 
can  be  written. 

If  the  vectors  are  drawn  carefully  to  scale  the  value  of  S  can 
be  found  from  the  vector  polygon.  The  method  of  construc- 
tion can  be  seen  from  Fig.  65,  §  80.  Start  at  m  and  lay  off  the 
vectors  in  order.  The  closing  side  S  is  the  resultant. 

This  method  of  solution  is  called  the  graphic  method. 


EQUILIBRIUM  OF  PARTICLES 


135 


A  third  method,  called  the  geometric  method,  is  as  follows: 
Draw  the  diagram  and  solve  the  problem  by  methods  of  Chap- 
ter VIII.  See  Figs.  72,  73. 

First  find  angle  a  from  the  known  directions  of  R  and  P. 
Then  AC  can  be  found.  Having  AC,  determine  angle  BAG 
and  angle  CAO.  Now  OC  and  angle  /3  can  be  found. 

Let  the  student  carry  out  the  work. 

It  is  strongly  recommended  that  every  problem  be  solved  by 
two  of  the  three  methods.  The  graphic  method  may  very  well 
be  used  as  one  in  each  case. 

II.  Consider  the  case  of  a  trap  door  as  shown  in  the  figure. 
It  is  desired  to  find  the  pull  (tension)  in  the  rope  and.  the  hinge 
reaction  in  the  position  given  in  Fig.  73. 


—Fa 


FIG.  73. 


FIG.  74. 


It  is  seen  at  first  that  the  weight  of  the  door  regarded  as 
applied  at  its  center  of  gravity,  G,  tends  to  produce  clockwise 
rotation  about  A,  the  hinge.  Further  the  rope  applied  at  B 
tends  to  produce  counterclockwise  rotation  about  A. 

The  measure  of  tendency  to  produce  rotation  by  a  force  is 
the  moment  of  the  force.  For  equilibrium  of  the  trap  door  the 
sum  of  the  moments  about  A  must  equal  zero.  That  is,  the 


136      POLAR  COORDINATES,  COMPLEX  NUMBERS,  VECTORS 

two  moments  must  be  equal  but  in  opposite  directions,  that  is, 
of  opposite  sign.     These  two  moments  must  be  calculated. 

The  moment  due  to  the  pull,  T,  in  the  rope,  is  equal  in  mag- 
nitude to 

M  =  T  -  AB  •  sin  DEC  (a  magnitude  of  the  vector  product) 
=  T-AB  sin  CBA 

=  T  •  10  •  sin  70°  =  9.39  T  (counter  clockwise). 
The  moment  of  the  weight  of  the  door  is 
Mz  =  w  -  AGsinCAG  (magnitude  of  vector  product) 
-  100  •  5  .  sin  60° 
=  433  (clockwise). 

/.     9.39  T  =  433 
and  T  =  46.1  Ibs.,  the  tension  in  the  rope. 

To  find  the  hinge  reaction  at  A,  we  find  its  horizontal  and 
vertical  components.  By  taking  horizontal  components  of  all 
forces,  calling  the  horizontal  component  H, 

T co&EBC  +  H  -  cos 0°  =  0, 
46.1  cos  140°  +  H  =  0, 

whence  H  =  35.4. 

By  taking  vertical  components, 

T  sin  EEC  +  V  sin  90°  +  W  sin  270  =  0, 
46.1  sin  140°  +  V  -  100  =?  0, 

whence  V  =  70.4. 

Now  if  P  is  the  hinge  reaction 

P  =  V(70.4)2+  (35.4)2  =  79.1,  approximately, 

704 
tan  6  =  ^2  =  2,  nearly, 

0  =  63°  25',  approximately. 

Thus   the   problem   is    completely 
solved. 

1.  The  forces  shown  in  the  dia- 
gram act  on  a  particle.  Calculate 
the  magnitude  and  direction  of  the  Ex.  1. 

resultant. 

Note.  —  Take  horizontal  and  vertical  components  as  in  1. 


EQUILIBRIUM  OF  PARTICLES 


137 


2.  Find  the  force  P  required 
to  hold  the  lever  AB  in  position 
as  shown  in  the  figure. 


-:*- 


51— \ 


Ex.  2. 


500 


3.  Find  P  so  that  the  load 
at  B  will  be  sustained,  CAB 
being  a  bent  lever  with  fulcrum 
at  A. 


Ex.  3. 


4.  Find  a  force  P  making 
an  angle  50°  with  OX  so  that 
the  system  will  be  held  in 
equilibrium.  Determine  also 
the  unknown  angle,  6. 


Ex.  4. 


5.  Three  men  carry  a  heavy 
uniform  bar  weighing  450  Ibs., 
one  man  at  one  end,  the  other 
two  with  a  short  stick  some  — 
distance  from  the  other  end. 
If  the  length  of  the  bar  is  I, 
find  how  far  from  the  end  the 
two  must  lift  so  all  lift  the  same  amount. 


Bar 


450 

Ex.  5. 


6.   Find  the  supporting    ^ 
forces  at  the  ends  of  the   I 
horizontal  bar  loaded  as 
shown  in  the  figure. 


100 


-9-- 


300 


Ex.  6. 


138      POLAR  COORDINATES,  COMPLEX  NUMBERS.  VECTORS 


7.  Find   the    stress   in 
the    rope    of    the    crane 
shown  in  the  figure.    Find 
the    forces    Vi,    Vz,    con- 
sidering ABC  as  a  single 
rigid  body  acted  upon  by 
Vi,  V2  and  2000  Ibs. 

Note.  —  Start  by  taking 
resolutions  at  (7,  then  mo- 
ments at  A. 

8.  Three  boys  pull  on 
three  ropes  tied  to  a  ring 
as   shown  in   the    figure. 
Determine    the    pulls    in 
ropes  AB,  AC.     (Resolu- 
tions.) 


9.  Determine  the 
forces  in  the  frame 
as  shown  in  the  fig- 
ure. 


10.  What   pull  on  the 
rope  is  necessary  to  hold 
the  boom  in  position  with 
the  load  as  indicated  in 
the  figure. 

11.  Find  the  resultant 
of  two  forces  of  500  Ibs. 
each,  one  acting  due  north 
and  the  other  N.  60°  E. 


Ex.  7. 


Ex.  9. 


Rope 


Ex.  10. 


CHAPTER  X 
EQUATIONS 

84.  Let/  (x)  be  defined  by  the  expression: 

/  (x)  =  a&n  +  aiz"-'  +  azxn~2  -f  •  •  •  +  an-ix  +  an, 

where  n  is  an  integer  and  the  a's  are  known  constants.     A 
function  of  this  kind  is  called  an  integral  function  of  the  variable 
x,  of  degree  n. 
Theorem  I.  —  If  r  is  a  root  of  the  equation: 

/(*)-0, 

then  /  (x)  is  exactly  divisible  by  x  —  r. 

For  divide/  (x)  by  (x  —  r)  by  the  ordinary  method,  continuing 
the  process  until  a  remainder  not  containing  x  is  obtained, 
the  result  can  be  represented  as 

(1)  f(x-)  =  (x-r)q(x)+R, 

where  q  (x}  denotes  the  quotient  and  R  the  remainder.  If  r 
is  a  root  of/  (x)  =  0,  the  left  side  vanishes  for  x  =  r.  The  first 
term  on  the  right  also  vanishes  for  x  =  r.  Then  (1)  becomes 

(2)  0  =  0  +  R 

whence  R  =  0.    Therefore,  the  division  is  exact  and 

(3)  f(x)  =  (x-r)q(x). 

85.  Assumption.  —  Every  integral  equation  *  has  at  least 
one  root. 

Theorem  n.  —  Every  integral  equation  of  degree  n  has 
n  roots.  Write  the  equation  as 

(4)  /(*)=0. 

*  That  is  an  integral  function  equated  to  zero. 
139 


140  EQUATIONS 

This  equation  has  a  root,  say  r\.    By  Theorem  I: 

(5)  /(*)  =  (« -r,)/!  (a?)  =0, 
where  f\  (x)  denotes  the  quotient.    Now  as  above 

(6)  /i(*)=0. 

If  /i  (x)  is  not  a  constant  it  is  an  integral  function  of  x  of 
degree  n  —  1,  and  Eq.  6  has  a  root,  say  r2.  Hence 

(7)  /i(*)  =  (*-*)/»(*). 

Continue  this  process  until  a  quotient  fn  (x)  not  containing 
x  is  obtained.  No  more  divisions  can  be  carried  out  and  no 
more  roots  exist.  The  result  is  that  /  (x)  has  been  broken  up 
into  factors  as  shown  in  the  equation, 

(8)  f(x)  =aQ(x-  n)  (x  -  r2)  .  .  .  (x  -  r«)  =  0. 

The  values  r\,  r2,  .  .  .  ,  rn  are  the  only  values  of  x  which  satisfy 
this  equation  and  consequently  the  only  roots  of  equation  (4). 

Incidentally  equation  (8)  shows  how  to  form  an  equation 
that  shall  have  given  roots. 

86.  Theorem  HI.  —  If  the  equation  /  (x)  =  0  has  more  than 
n  roots  it  is  an  identity  and  has  infinitely  many  roots  and  every 
coefficient  is  zero. 

Consider, 

(9)  /  (x)  =  aozn  +  aixn~l  +  '•••+  ctn-is  +  an  =  0 

and  suppose  it  has  more  than  n  roots.  If  every  a  is  not  zero 
the  terms  whose  coefficients  do  not  vanish  will  form  an  equation 
of  degree  not  higher  than  n  which  can  have  not  more  than  n 
roots.  But  this  contradicts  the  hypothesis.  Hence  the  theorem 
is  true. 

This  theorem  has  important  uses  in  mathematics,  some  of 
which  will  appear  in  the  sequel. 

Determine  which  are  identities  and  which  are  not. 

1.]  (x  —  a)2  =  z2  —  2  ax  +  a2,  expand  left  side,  transpose 
and  collect  terms. 

2.  x2  —  4  =  0.  For  how  many  values  of  a;  is  this  equation 
satisfied? 


ZERO  AND  INFINITE  ROOTS  OF  INTEGRAL  EQUATIONS     141 

«.2   Q 

3.  — r- ~  —  x  —  3.     Clear  of  fractions. 
x  +  3 

4.  r5  -  3  x2  -  x  =  0. 
5. 


1 -X  ' 1 -3 

6.  Is  2  a  root  of  x2  +  4  z  +  4  =  0?     Is  -2  a  root? 

7.  Is  1  a  root  of  z3  -  3  z2  +  3  x  +  1  =  0?     Is  2  a  rootl 

87.  Theorem  IV.  —  If  f  (x)    be   divided   by    (x  —  r),    the 
remainder  will  be  /  (r). 

For  write  /  (x)  =  (x  —  r)  q  (x)  +  R. 

Now  put  x  =  r        f  (r)  =  (r  —  r)  q  (r)  +  R- 
Sincer-r  =  0,  R=f(r). 

This  theorem  furnishes  a  convenient  method  of  calculating 
f  (r).  To  shorten  the  work  it  will  be  desirable  to  learn  an 
abbreviated  method  of  performing  the  division  by  x  —  r.  This 
will  be  given  later. 

88.  Zero  and  infinite  roots  of  integral  equations.  —  Con- 
sider 

(1)  f(x)  =  aozn  +  ajx"-1  +  •  •  •  +  a*-\x  +  an  -  0. 

If  an  =  an-i  =   •  -  •   =  an-jt+i  =  0,  k  roots  of  (1)  are  zero. 
Under  this  hypothesis  (1)  may  be  written, 

(2)  a&n  +  Oix""1  +  •  •  •  +  an-ix  +  an 

=  xk  (a&n-k  +  atf1-*-1  +  •  •  •  +  fln-t)  =  0. 

It  is  seen  now  that  k  roots  are  zero  since  xk  =  (x  —  0)fc  is  a 
factor  of  the  expression  constituting  the  left  member. 

Now  substitute  x  =  -  in  (1)  and  obtain,  after  multiplying 

by  yn, 

(3)  anyn  +  an-iyn~l  +  •  •  •  +  a#  +  a<>  =  0. 

If  in  (3)  OQ  =  ai  =  02  =   •  •  •   =  ak-i  =  0,  k  roots  are  zero,  as 

shown  above.     But  by  virtue  of  the  relation,  x  =  -,  x  =  oo 

y 

when  y  —  0.    Therefore,  there  is  an  infinite  root  of  (1)  for  each 


142  EQUATIONS 

zero  root  of  (3).    Hence  when  the  first  k  coefficients  of  (1) 
approach  zero,  k  of  its  roots  become  infinitely  large. 

Zero  roots  often  occur  in  practical  work,  and  infinite  roots 
have  important  meanings  in  certain  types  of  problems. 

1.  What  is  the  value  of  x  in  xy  =  k,  when  y  =  0?     If  this 
equation  is  regarded  as  the  equation  of  a  curve  what  is  the 
interpretation  for  y  =  0? 

2.  What  is  the  value  of  y  =  x2  —  2  x  +  1,  when  x  =  1? 
Interpret  this  result  as  in  Ex.  1. 

89.  Synthetic  division  is  a  shortening  of  the  process  of  long 
division  when  the  divisor  is  a  binomial  of  the  first  degree  in  the 
variable.  Consider  the  example: 

x3  -5xz  \xz  +  x  -2 

x2  —  7  x 


-2s  +  10 
If  the  cancelled  terms  are  omitted  the  work  appears  as 


-5x2  \x*  +  x  -  2 

x2  -7x 
-5x 


+  10 

By  pushing  up  the  remainders  into  a  line  under  the  dividend 
the  work  appears  as 


-  5  x2  -  5  x  +  10 
-  2  x        0 

The  x's  may  be  omitted,  using  coefficients  only.     Thus 
1-4-7  +  10    |-5 
-5-5  +  10        |l  +  l-2 
1-2       0 


THEOREM  V  143 

It  is  noticed  that  the  last  form  gives  all  the  information  given 
in  the  first  form  of  division.  If  the  coefficient  of  the  first  term 
of  the  dividend  is  brought  down,  then  the  partial  remainders 
hi  order  are  the  coefficients  of  the  quotient  and  the  final  re- 
mainder is  zero. 

In  applying  this  method  it  is  desirable  to  use  addition  instead 
of  subtraction  during  the  process.  This  is  done  by  changing 
the  sign  of  the  second  term  of  the  divisor,  that  is  change  —5 
to  5.  Thus, 

1-4-7  +  10    [5 
+  5  +  5-10 

1+1-2        0 

90.  Theorem  V.  —  If  /  (x)  is  exactly  divisible  by  x  —  r,  r  is 
a  root  of  the  equation  /  (x)  —  0. 

This  theorem  is  proved  by  means  of  the  equation  used  to 
prove  Theorem  I.  For  if  R  =  0, 

f(x)  =  (x  -r}q(x'). 

Substituting  r  for  x  causes  the  right  side  to  vanish,  and  hence 
the  left  side  also.  Therefore,  r  is  a  root  of  /  (x)  =  0,  by  defini- 
tion of  root. 

It  is  now  easy  to  use  Theorem  V  and  synthetic  division  to 
determine  whether  or  not  any  given  number  is  a  root  of  a  given 
integral  equation.  If  a  number  should  not  be  a  root  it  is  useful 
to  know  that  the  final  remainder  R  is  the  value  of  /  (r). 

In  applying  synthetic  division,  if  any  term  in  £  is  missing 
its  place  must  be  filled  with  a  zero. 

Solve  by  synthetic  division: 

1.  Is  2  a  root  of  x3  -  x*  -  3  x  -  3  =  0? 

2.  Is  1  a  root  of  z3  -  z2  +  3  Z  -  3  =  0? 

3.  Is  x  -  3  a  factor  of  3  z4  -  11  x3  +  5  x2  +  3  x  =  0. 

4.  Factor  6  x2  +  19  x  +  10. 

5.  Factor  x3  +  5  xz  +  2  x  +  10. 

6.  Factor  x*-2x*  -  8  z  -  16  (supply  term  in  z2  by  0). 


144  EQUATIONS 

7.  Find  any  integral  root  of  x3  —  25  xz  +  8  x  —  16  =  0. 

8.  Find  any  integral  root  of  2  z3  -  3  z2  —  3  z  +  2  =  0. 

9.  Find  any  integral  root,  of  z4  +  2  z3  —  5  z2  —  4  z  +  6. 

91.  Solution  of  numerical  equations  in  one  unknown  of 
any  degree.  —  No  general  simple  rule  can  be  given  for  finding 
the  roots  of  equations  of  degree  higher  than  the  second.  It  is 
possible  to  solve  equations  of  the  third  and  fourth  degrees  by 
formulas,  but  the  application  is  in  general  not  easy.  For 
practical  purposes  some  method  of  approximation  is  most 
useful.  Such  a  method  can  be  easily  constructed  from  the 
preceding  theorems. 

For  values  of  x  not  roots  of  /  (x)  =  0,  it  is  obvious  that  the 
result  of  substituting  such  values  in  /  (x)  would  give  positive 
or  negative  values  of  the  function  instead  of  zero. 

For  values  of  x  near  r,  where  r  is  a  root  of  /  (z)  =0  and  h 
a  small  positive  number,  one  of  the  following  relations  must,  in 
general,  hold: 

(a)    /(r-*)</(r)=0</(r  +  ft), 
or  (6)    /(r-/0>/(r)=0>/(r  +  A), 

or  (c)     /(r-ft)</(r)  =0>/(r  +  A), 

or  (d)    /(r-*)>/(r)~0</(r  +  *)l 

or  (e)     /(r-ft)=/(r)=0=/(r  +  ft). 

Conditions  (a)  and  (6)  are  the  ones  which  will  now  be  con- 
sidered, being  the  most  commonly  occurring.  These  can  be 
used  to  discover  the  position  of  the  real  roots  of  an  equation  of 
any  form. 

Consider  first  an  integral  equation, 

/(z)  =  z3-4z2-2z  +  8  =  0. 

Note.—  What  follows  illustrates  a  method.  In  practice  one 
woif  d  first  determine  whether  integral  factors  of  8  are  roots. 

Try,  by  synthetic  division,*  in  succession  the  values  —3,  —2, 
—  1,  0,  1,  2,  3,  etc.: 

*  Synthetic  division  is  here  only  a  convenient  method  of  substituting 
values  of  x  in  the  equation. 


SOLUTION  OF  NUMERICAL  EQUATIONS  145 


1-4-    24-    8    1-3 
-3  +  21-57 
-7  +  19-49 


-2  +  12-20  x  =  -3,  R  =/(-3)  =  -49 

x=  -2,R=f(-2)  =  -28 
x  =  -1,  #=/(-!)  =  +5 
By    condition     (a)     a    root    lies 
between  —2  and  —1. 


=  +8 
x=  l,R=f(l)  =  +3 
x  =  2,R=f(2)  =  -4 
By    condition    (6),    a    root    lies 
between  1  and  2. 


1-4-    2+    8 

2-    4-12 

-2-    6-    4 


=  -7 

'r  =  4    7?  =  f  61^  =  0 

*(•  *T  ,      J  i  I      \~tj  vl  i 

4isaroot. 

0-2        0 

Bringing  down  the  first  coefficient  the  quotient  is  x2  —  2. 
Now  having  found  one  root,  4,  exactly  the  equation  can  be 
written 

x3  -  4  x2  -  2  x  +  8  =  (x  -  4)  (x2  -  2)  =  0 

and  the  remaining  roots  are  easily  found  from  the  equation 

x2  -  2  =  0. 

But  to  illustrate  the  method,  when  an  exact  root  is  not 
found  the  fact  that  4  is  a  root  will  be  ignored  and  we  shall 
proceed  to  find  the  root  that  lies  between  —2  and  —1. 


146  EQUATIONS 

Take  a  value  of  x  midway  between  —2  and  —  1,  that  is  —1.5, 

1-4    -2      +8         | -1.5  s= -1.5.  fl=/(-1.5)= -1.375 

-  1.5  +  8.25  -  9.375  By  condition  (a)  a  root  lies 

-  5.5  +  6.25  -  1.375  between  -1.5  and  -1. 

Now  take  the  value  midway  between  —1.5  and  —  1.  Since 
this  value  involves  three  figures,  it  will  be  better  to  take  -a 
near  value  in  two  figures  as  —1.2  or  —1.3.  Again,  since  the 
value  of  R  for  x  =  —1.5  is  smaller,  numerically,  than  the  value 
of  R  for  x  =  —1,  we  will  risk  choosing  —1.3. 

1-4     -2      +8        1-1.3 


689-636  *=  -1.3,  fl  =  /  (-1.3)  =  1.64. 

l.O  T^  O.OW    —  O.OD  T-»         j.  1  -I    r  1  1    n 

—  _  0   i    .  -_   ,   .,  -  .  Root  between  —1.5  and  —1.3. 

—  o.o  •+•  4.o9  +  1.64 

Now  take  —1.4,  midway  between  —1.5  and  —1.3. 

1-4     -2      +8        LlL*    z=  -1.4,  fl=/(-1.4)  =  +0.216. 

-  1.4  +  7.56  -  7.784  Root  between  _IA  and  _L5> 

-  5.4  +  5.56  +  0.216 

Since  /(  —  1.4)  is  much  smaller  (numerically)  than  /(—  1.5) 
we  will  try  values  quite  near  —1.4.     Take  —1.41. 

1  -4       -2         +8 

-  1.41  +  7.628  -  7.935 

-  5.41  +  4.628  +  0.065 
Take  -1.42, 

1_4       _2        +8 


-1.42  +  7.696-8.088  Root^between  -1.41    and 

-  5.42  +  5.696  -  0.088 

The  root  is,  therefore,  1.41  correct  to  three  figures.  The  nearly 
equal  values  of  R  for  —1.41  and  —1.42  would  suggest  that  the 
next  figure  is  near  5  and  a  continuation  of  the  work  will  show 
that  the  next  figure  of  the  root  is  4  and  the  root  is  —1.414 
correct  to  four  figures.  This  value  is  sufficiently  exact  for  all 
ordinary  purposes. 

The  above  process  is  rather  laborious.     Practice  will  enable 
one  to  obtain  three  or  four  figures  of  a  root  quite  readily.    None 


r  , 

~"'        /(  > 


SOLUTION  OF  NUMERICAL  EQUATIONS  147 

of  the  current  methods  of  finding  irrational  roots  are  much 
shorter,  if  any.  Some  are  longer  and  require  more  theoretical 
knowledge. 

1.  Find  the  roots  of  z4  +  x5  -  7  x*  -  5  x  +  10  =  0. 

2.  Find  the  roots  of 

.    28z4  +  239z3  +  1020z2  +  813z-140  =  0. 

3.  Find  the  root  of  Ex.  13,  following  34,  by  the  above  method. 

Consider  another  type  of  equation, 

x  —  sin  x  —  1  =0. 

Synthetic  division  cannot  be  employed  here  as  a  method  of 
substituting  values  of  x,  for  the  reason  that  this  equation,  as 
will  be  shown  later,  has  an  infinite  number  of  terms  when  we 
expand  it  into  an  integral  equation.  Such  an  equation  is  called 
transcendental.  In  this  case  the  angle,  x,  in  the  first  term 
must  be  given  in  radians.  The  values  of  sin  x  are  to  be  taken 
from  a  table  of  natural  sines. 

Try  x  =  2  radians  =  114°  35',  nearly. 
2  -  0.9095  -  1  =  +0.0905,  x  =  2  rad.,  R=f(2)  =  0.0905. 

If  x  =  1.9  radians  =  108°  51' 


19    09463-1-  -00463  --»    «=/  (1-9)  =  -0.0463. 

l.y  —  U.y^rOO  —  1  —  —  U.vrrUO,    I   -r>        j.   I.  ir»  in  i- 

J  Root  between  1.9  and  2  radians. 

If  x  =  1.95  radians 
1.95  -  0.9291  -  1  =  0.0209,  x  =  1.95  rad.,  R  =  f  (1.95)  =  0.0209. 

If  x  =  1.93  radians 
1.93-0.9362-1=  -0.0062,  x  =  1.93  rad.,  #=/(!.  93)  =  -0.0062. 

This  value  is  correct  within  an  error  of  less  than  angle  of  10'. 
Further  trials  will  give  more  accurate  results. 
Solve  by  repeated  substitution  : 

1.  x  =  tanx. 

2.  x  =  3  sin  a;  —  1. 

3.  3  x  +  logio  x  =  5. 

4.  sinz  +  cosx  =  1.4. 


148  EQUATIONS 

Note.  —  In  some  cases  it  may  be  useful  to  construct  the 
graph  of  the  equation  before  attempting  the  solution.  The 
intercept  of  the  graph  on  the  axis  of  abscissas  will  furnish  a 
guide  in  selecting  trial  values  for  the  root. 

5.  The  distance  from  the  earth  of  a  body  projected  vertically 
upward  is  given  by  the  formula 

s  =  v0t-  16.1 12, 

where  v0  is  the  velocity  of  projection  upward  in  feet  per  second 
and  t  is  the  time  from  starting,  in  seconds  and  s  is  the  distance 
in  feet.  If  a  body  is  projected  upward  with  a  velocity  of 
80'/sec.,  in  how  many  seconds  from  starting  will  it  be  30'  from 
the  earth?  (Two  values  of  t.} 

6.  A  stone  is  dropped  into  a  well.     It  is  heard  to  strike  the 
bottom  after  5  sec.     Having  given  s  =  16.1  t2  and  the  velocity 
of  sound  =  1150'/sec.,  determine  the  depth  of  the  well. 

7.  The  volume  of  a  cubical  box  is  diminished  1200  cu.  in.  by 
putting  in  it  1/2  in.  lining  on  all  faces.     Find  the  original 
dimensions  of  the  box. 

8.  The  compound  amount  of  $3000  at  x  per  cent,  for  5  yrs., 


is  $3500 


(x  Y 
1  +  155]  • 

9.  Find  the  length  of  a  chord  of  a  circle  that  cuts  off  \  its 
area  if  the  radius  is  1. 

10.  Find  the  x-mtercepts  of  the  curve  whose  equation  is 

y  =  2xs  -z2-6z  +  3. 

Note.  —  Write  the  function  equal  to  zero  and  solve  the 
equation  for  its  roots. 

11.  How  deep  will  a  sphere  of  wood  1'  in  diameter  sink   in 
water  if  the  density  of  the  wood  is  0.7  that  of  water,  given 
the  volume  of  a  spherical  sector  equals  the  area  of  its  spherical 
surface  times  ^  its  radius  and  the  volume  of  a  cone  equals  |  its 
base  times  its  altitude. 

12.  Find  the  real  fifth  root  of  15. 


PARTICULAR  CASE  OF  QUADRATIC  EQUATION       149 

Note.  —  x5  —  15  =  0.  Solve  this  equation  in  the  regular 
way. 

13.  In  the  solution  of  a  certain  problem  in  mechanics  the 
result  depended  on  solving  the  equation 

cos3  0  +  0.8  cos2  6  -  0.02  cos  0  -  0.393  =  0. 

Determine  one  root  to  three  figures,     (cos  0  as  unknown.) 

92.  Particular  case  of  quadratic  equation.  —  The  frequent 
occurrence  of  equations  of  the  second  degree  (quadratic  equa- 
tions) makes  it  desirable  to  give  them  some  special  treatment. 
All  the  theorems  regarding  integral  equations,  given  in  this 
chapter  so  far,  hold  for  quadratic  equations. 

By  Theorems  II,  III  a  quadratic  equation  has  two  and  only 
two  roots.  It  can,  therefore,  be  written  in  the  form 

(1)  k(x-  n)  (x  -  r,)  =  0, 

where  k  is  constant  and  rt,  r2  are  the  roots.     Equation  (1)  can 
be  written 

(2)  kx*  -k(ri  +  r2)  x  +  km  =  0 
or  xz  —  (TI  +  r2)  x  +  rir2       =  0. 
The  equation  (2)  is  of  the  form 

(3)  az2  +  bx  +  c  =  0 

or  z2  +  -z  +  -  =  0. 

a         a 

Comparing  the  left-hand  members  of  (2)  and  (3)  it  is  easily  seen 
that  if  they  represent  the  same  equation 

b  c 

TI  +  TZ  = and    rir2  =  — 

a  a 

The  formula  given  in  10  may  be  obtained  as  follows: 
Given 

axz  +  bx  +  c  =  0. 
Dividing  by  a, 

.  6  c 

x*  +  -x  = 

a  a 


150  EQUATIONS 

bz 

Adding  -. — 5  to  both  members. 
4  a2 

,.2_L&      ,_&!_       *>*       c_ 

*"«**  1~4a2~4a2      a~       4a2 

Taking  the  square  root  of  the  first  and  last  members, 

_6_ 

" 


or  x  = 


2a 

-6=b\/62-4ac 
2a 


The  two  roots  of  the  quadratic  will  be  equal  to  each  other  if 
62  —  4  ac  *  =  0.  In  this  case  each  root  is  —6/2  a. 

What  will  be  the  nature  of  the  roots  if  62  —  4  ac  <  0? 

What  will  be  the  nature  of  the  roots  if  6  —  4ac>0? 

By  calculating  the  expression  62  —  4  ac  it  is  possible  to  know 
the  nature  of  the  roots  of  the  quadratic  equation  without 
actually  calculating  them.  Thus,  if 

xz  +  x  +  1  =  0, 
then, 

a  =  l,    6  =  1,    c  =  l,    62-4oc=-3. 

Therefore  the  roots  are  complex  and  unequal. 

Determine  the  nature  of  the  roots  of  the  following: 

1.  4z2  +  8z  +  4  =  4.  2.   3z2-6z  +  3  =  0. 

3.      x2  -  7  x  +  16  =  0.  4.   9  x2  -  12  x  -  6  =  0. 

5.  If  c  =  0,  one  root  of  the  quadratic  is  0,  88.  If  6  =  0,  the 
roots  are  equal  numerically  but  of  opposite  signs.  They  are 

given  by  ±  \ 

*        Of 

93.  Equations  of  higher  degree  than  the  second  may  some- 
times be  written  in  quadratic  form.  The  following  is  such  an 
equation, 

2x6  -5^  + 2  =  0. 
*  This  expression  is  called  the  discriminant  of  the  quadratic. 


EQUATIONS  151 

First  consider  x5  as  the  unknown,  then  the  equation  is  a  quad- 
ratic.    Solving  by  the  formula  of  the  preceding  section 

5  ±  V25  -  16  .  1 

x3  =  -  £  -  =  2,  and  -^ 

The  original  equation  can  now  be  written  in  the  form 

(z3  -  2)  (tf-  |)  =  0. 
It  is  necessary  now  to  solve  the  two  equations 

a?  -  2  =  0,    xs  -  \  =  0. 
Whence, 

(x  -  #2)  (xz  +  ^2x  +  ^4)  =  0; 


From  the  second  factor  of  the  first  equation,  by  the  formula 

_^±V^-4^4 
(a)  *  =  -  -3- 

The  approximate  values  of  the  radicals  should  be  substituted 

and  the  values  of  x  expressed  in  usable  form.     From  the  first 

factor  of  the  same  equation 

(6)  x  =  #2. 

From  the  second  factor  of  the  second  equation, 


From  the  first  factor  of  the  same  equation, 

(d)  x  =  V\. 

There  are,  altogether,  six  roots;  these  should  be  expressed  as 
decimals  to  four  figures.  The  solution  of  this  problem  should 
be  carefully  mastered  as  a  typical  case  of  higher  equations  in 
quadratic  form.  Use  table  I  for  evaluating  radicals. 


2.  z  +  2a;  +  l  =  0. 

3.  a?  -3s  +  2  Vs2-3x  +  2  =  l. 

4.  Vs  —  1  -  5  -Vx  —  1  =  -6. 


CHAPTER  XI 
THE  LINEAR  FUNCTION  AND  THE  STRAIGHT  LINE 

94.  The  most  general  function  of  the  first  degree  in  one 
variable  is  of  the  form: 

(1)  f(x)  =  mx  +  b, 

where  m  and  b  are  constants.     The  most  general  equation  of 
the  first  degree  in  two  variables  is  of  the  form: 

(2)  Ax  +  By  +  C  =  0, 

where  .A,  B  and  C  are  constants.    Solving  (2)  for  yt 

(3)  y=-*x-C 

which  is  of  the  form 

(4)  y  =  mx  +  6. 

Equation  (4)  is  of  the  same  form  as  (1)  except  that  y  is  written 

for  /  (x).     Equation  (3)  shows  that  (2)  can  be  written  in  the 

form  of  (4)  .     Since  (3)  is  obtained  from  (2)  by  transposing  and 

dividing  by  a  constant,   it  is  virtually  the  same  equation. 

Since  (4)  is  of  the  same  form  as  (3),  any  conclusions  drawn 

from  (4)  will  be  valid  for  (3)  and  consequently  for  (2). 

96.  Theorem.  —  The  graph  of  any  equation  of  the  first 
degree  in  two  variables  is  a  straight  line. 

Using  equation  (4),  let  (xi,  7/1)  be  any  fixed  point  on  the  locus 
or  graph.  Let  (xz,  y%)  be  any  other  point  on  the  graph,  chosen 
arbitrarily.  Then  by  34, 

yi  =  mxi  +  6, 
y2  =  mxz  +  b. 
Subtracting  and  solving  for  m, 


152 


FAMILIES  OF  LINES 


153 


Since  m  is  constant  by  hypothesis,  Eq.  (4),  94,  and  since  Xz,  y% 
is  any  point  different  from  (x\,  yi)  it  is  evident  the  slopes  of  the 
segments  joining  any  two  points  of  the  locus  are  equal.  This 
can  be  true  only  if  the  locus  is  a  straight  line. 

If  6  is  the  angle  be- 
tween the  a>axis  and 
the  line,  the  slope  of 
the  line  is 


FIG.  75. 


where  (x\,  yi),  (x2,  y2) 
are  any  two  points  on 
the  line.  The  angle, 
6,  is  called  the  inclin- 
ation of  the  line  with 
the  axis. 

Note  that  the  slope  is  the  coefficient  of  x  in  the  form  (4),  94. 

1.   Determine  the  slope,  intercepts  and  draw  the  lines  repre- 
sented by  the  following  equations: 

a.  2 x  -{-  3y  —  1=0. 

b.  x  —  y  =  14. 

c.  x  +  y  =  —14. 


FIG.  76. 


96.  If  in  Equation  (4), 
m  remains  fixed  while  b 
varies,  there  results  an 
infinite  number  of  equa- 
tions representing  an  in- 
finite number  of  parallel 
lines  covering  the  whole 


plane.     A  set  of  lines  having  a  common  property  is  called  a 
family  of  lines. 

If  b  remains  fixed  while  m  varies  there  results  a  family  of  lines 
passing  through  tne  same  point  (0,  6),  on  the  y-axis. 


154     THE  LINEAR  FUNCTION  AND  THE  STRAIGHT  LINE 

When  a  coefficient,  which  is  ordinarily  constant  in  an  equa- 
tion, is  made  to  vary  hi  the  manner  above  indicated,  it  is  called 
Y       /  a  parameter.    The  family 

of  lines  obtained  by  means 
of  one  such  parameter  is 
called  a  one-parameter  f  am- 

Draw  several  lines  from 
each  of  the  equations  below 
by  assigning  different  values 
to  the  parameter. 

Fir1    77 

Give  &  different  values  and 
draw  a  line  for  each  value  of  6. 

y  =  mx  +  4. 
Give  m  different  values  and  draw  a  line  for  each. 

97.   Converse   theorem.  —  Every   straight   line   is   repre- 
sented by  an  equation  of  the  first  degree  in  two  variables. 


' 

x^lr 

,,;>  " 

i* 

*£Z 

\l 

Af^W~ 

^^  ^ 

T   1 

ii 

^|      XZ~X1     1 

1 

^s"^ 

f<  1-  OKEj-- 

-H 

^s^ 

i              i 

1     Y 

.^           O 

FIG.  78. 


Let  (xi,  ?/i)  and  fa,  f/2)  be  any  two  given  points  on  the  line. 
Let  (x,  y)  be  any  third  point  on  the  line.    The  slope  of  PI  P  is 


x  —  x\ 


THEOREMS  155 

The  slope  of  PiP2  is 

3/2  -yi 


Since  the  three  points  are  on  a  straight  line  the  two  slopes 
just  written  are  equal.     Hence 

y-yi  =  3/2  -  3/1 
x  —  Xi      Xz  —  Xi 

(5)     Or  y-y1=yl(x-Xl}, 


This  equation  is  of  the  first  degree  in  the  variables  x  and  y 
Hence  the  theorem  is  true. 

1.   Find  the  equation  of  the  straight  line  through  (—2,  1) 
and  (5,  —4).    Using  (5)  above 


or  y-l  =  -f  (z  +  2). 

7?/  +  5z  +  3  =  0. 

2.  Find  the  equation  of  the  line  through  (1,  3)  and  (2,  5). 

3.  The  slope  of  a  line  is  2.     It  passes  through  the  point  (3,  7). 
Find  its  equation. 

Note.  —  Use  equation  (5),  noting  that 

y*  ~yi  =  m  =  slope. 

Xz-Xi 

4.  The  slope  of  a  line  is  m.    It  passes  through  the  point 
fa,  3/1).     Find  its  equation. 

5.  Is  the  point  (6,  4)  on  the  line  y  =  4  x  —  2  ? 

6.  Are  the  points  (3,  1),  (4,  3),  (6,  8)  on  the  same  straight 
line? 

7.  Find  the  equation  of  the  line  through  (5,  7)  and  having  a 
slope  of  —  £. 

8.  Find  the  equation  of  the  line  through  (1,  2),  (  —  1,  6). 

9.  Find  the  equation  of  the  line  through  (3,  —5)  and  parallel 
to  the  line  3  x  -  4  y  +  2  =  0. 

10.  What  is  the  equation  of  the  line  parallel  to  the  z-axis  and 
distant  4  units  from  it?     (Two  solutions.) 


156     THE  LINEAR  FUNCTION  AND  THE  STRAIGHT  LINE 

98.   Let  (x,  y)  be  the  coordinates  of  any  point  P  on  the  line 
AB  and  p  =  Oe  the  distance  of  the  line  from  0.     Let  the  angle 
TOe  be  a.    Now  the  projections  of  x  and  y  along  Oe  are  such 
that 
(6)  x  cos  a  +  y  sin  a  =  p. 


FIG.  79. 

This  relation  holds  for  all  points  on  AB.  Therefore  (6)  is  the 
equation  of  the  line  AB.  Equation  (6)  is  called  the  normal  form 
of  the  equation  of  a  straight  line.  This  form  is  useful  in  prob- 
lems relating  to  the  distance  of  a  point  from  a  line. 

Let  A  'B'  be  a  line  parallel  to  AB,  and  let  (xi,  y\)  be  any  point 
on  A  'B'.    Then  if  pi  =  Oe', 
(a)  Xi  cos  a  +  y\  sin  a  =  pi  =  p  +  d, 

where  d  is  the  distance  of  (x\,  y\)  from  AB,  or  the  distance 
between  the  lines.     From  (a)  by  transposing, 
(&)  Xi  cos  a  +  yi  sin  a  —  p  =  d. 

The  left  side  of  (&)  is  exactly  what  (6)  becomes  if  p  is  transposed 
to  the  left  side  and  (xi,  yi)  substituted  for  (x,  y).  Hence  the 
distance  of  any  point  (x',  y'}  from  a  line  AB  may  be  found  by 
reducing  the  equation  of  AB  to  normal  form  and  transposing 
all  terms  to  the  left  side,  then  substituting  the  value  (xf,  y') 
for  (x,  y).  The  result  is  the  distance  of  the  point  (xf,  y')  from 


LINEAR  FORMS  157 

AB.  This  distance  is  to  be  considered  positive  if  the  point 
(zi,  7/1)  and  the  origin  are  on  opposite  sides  of  the  line  AB,  and 
negative  if  the  point  (zi,  «/i)  and  the  origin  are  on  the  same  side 
of  the  line  AB, 

From  the  triangle  SOT,  p  =  b  sin  a  =  a  cos  a,  and  a/b  = 
tan  a.     From  47  it  is  seen  that 

a  b 

sin  a  =      .          —  ,  cos  a  =      ,          =  - 
Va?  +  b2  Va2  +  62 

By  94,  Eq.  3,  since  a  and  b  are  the  intercepts,  if  the  equation  of 
a  line  is 

Ax  +  %  +  C  =  0, 
then 


a-  - 


Va2  -j-  62 
It  follows  that 

-C/B  -A 


cos  a  = 


-CIA 
sin  a  =  - 


'cys2     ± 
-C 


'A2  +  #2 

Thus  if  Ax  +  By  +  C  =  0  is  the  general  form  of  the  equation 
of  a  straight  line,  then 

A  B  -C 

(7)  /,.  .    ™*  +  ~ 


is  the  normal  form  of  the  equation  of  the  same  line.  It  is  cus- 
tomary to  choose  the  sign  of  the  radical  in  (7)  so  that  the  right 
member  of  the  equation  shall  be  positive. 

99.   The  different  forms  of  the  equation  of  the  straight  line 
in  common  use  may  be  summarized  as  follows: 

1.  Ax  +  By  +  C  =  0,  general  form. 

2.  y  =  mx  +  6,  slope-intercept  form. 

3-  y  —  yi  —  m(x  —  Xi),  one-point  slope  form  (see  Ex.  4,  97). 


158    THE  LINEAR  FUNCTION  AND  THE  STRAIGHT  LINE 

4.  y  —  yi  =  [(2/2  -  y\)/(xz  —  Zi)]  (x  —  Zi),  two-point  form. 

5.  x/a  +  y/b  =  1,  intercept  form. 

To  obtain  (5),  divide  (1)  by  —C  and  note  that  the  values  of 
a,  6  above  are  the  intercepts  of  the  line. 


6.  x  cos  a  +  y  sin  a  =  p  or 
A  B 

X 


-C 


Normal  form. 


These  forms  are  to  be  memorized.  Their  use  is  somewhat 
suggested  by  their  names.  In  a  problem,  careful  attention  to 
what  is  given  regarding  a  line  will  often  suggest  what  form  of 
the  equation  is  to  be  used. 

100.   To  find  the  distance  between  two  points,  having  given 
their  coordinates. 


FIG.  80. 
From  the  figure  it  is  seen  that  13, 


This  formula  is  true  for  points  in  all  positions.     Care  must  be 
used  regarding  the  signs  of  the  coordinates. 

101.  The  coordinates  of  a  point  that  divides  the  segment 
joining  two  points  in  a  given  ratio  can  be  found  as  follows: 
Let  AB  be  the  segment  and  P  the  point  of  division  and  r  the 


DIVISION  OF  A  SEGMENT 


159 


ratio  of  the  two  parts  into  which  AB  is  to  be  divided.     From 
the  similar  triangles  in  the  figure 

AP  =  AM 
PB~  PS 


=  r. 


FIG.  81. 
or  since  AM  =  x  —  Xi,  PS  =  x%  —  x  the  equation  becomes 

x  —  x\ 


—  x 


=  r. 


and 


In  a  similar  way, 


x  = 


y  = 


r  +  l 


r+l 


If  P  is  not  between  A  and  B  the  ratio  r  is  negative.  The 
line  AB  is  then  divided  externally. 

1.  Find  the  coordinates  of  the  points  of  trisection  of  the 
segment  joining  A(2,  1)  to  B(8,  —4). 

Note.  —  Since  there  are  two-  points  of  trisection  the  solution 
is  to  be  done  in  two  parts.  First  let  r  =  |,  then 


x  = 


=  4 


160     THE  LINEAR  FUNCTION  AND  THE  STRAIGHT  LINE 


and  similarly  for  y.     For  the  second  point  call  r  =  2  and  use 
the  same  formula.     Student  complete  the  solution. 

2.  Find  the  middle  point  of  the  segment  in  Ex.  1. 
Note.  —  Call  r  =  1  and  proceed  as  above. 

3.  Find  the  coordinates  of  the  point  nearest  (2,  1)  that 
divides  the  line  in  Ex.  1,  externally  in  the  ratio  of  £. 

Note.  —  Callr  =  -f. 

4.  Find  the  coordinates  of  the  point  that  divides  the  line  in 
Ex.  1,  externally  in  the  ratio  of  1  to  1.     What  is  the  geometric 
interpretation  ? 

102.   The  angle  between  two  lines  can  be  determined  from 
their  slopes.     It  is  evident  that  in  the  figure  0  =  02  —  0i. 


Hence 


tan  $  =  tan  (02  —  0i) 

tan  02  —  tan  0i 


1  +  tan  02  tan  6\      1  +  m\m^ 


FIG.  82. 

But  tan  02  and  tan  0i  are  the  slopes  of  A 'B'  and  AB  respectively. 
The  slopes  are  to  be  found  by  any  available  method  and  sub- 
stituted in  the  above  equation.  The  result  is  the  tangent  of 
the  angle  <£.  In  applying  this  rule  it  is  desirable  to  take  for  02 
the  larger  of  the  two  angles  02  and  0i  if  it  is  possible  to  determine 


MISCELLANEOUS  EXERCISES  161 

which  is  the  larger.  Then  0  is  the  angle  through  which  A  B 
must  revolve  in  the  positive  direction  (counter  clockwise)  to 
bring  it  into  parallelism  with  A'B'. 

1.  Find  the  angle  between  the  lines  2x  —  3  i/  -f-  4  =  0  and 
x  —  y  =  1. 

Note.  —  Find  the  slopes  from  the  equations  and  substitute 
the  slopes  in  the  above  formula. 

2.  Find  the  angles  of  the  triangles  whose  vertices  are  (1,  1); 
(6,8);   (7,  -3). 

Note.  —  From  the  coordinates  of  the  points  in  pairs  find  the 
slopes  of  the  lines  joining  them  and  proceed  as  in  Ex.  1. 

3.  What  relation  must  hold  between  tan  02  and  tan  0:  in  order 
that  A'B'  shall  be  parallel  to  AB1 

4.  Knowing  that  tan  90°  =  oo  ,  show  that  AB  will  be  per- 
pendicular to  A'B'  if  tan  02  •  tan  0i  =  —  1. 

5.  Show  that  the  figure  whose  vertices  are  (0,  1);  (2,  0); 
(5,  6)  ;  (3,  7)  is  a  rectangle. 

MISCELLANEOUS  EXERCISES 

1.   What  is  the  slope  of  each  of  the  following  lines? 
(a)    10y  +  3z  =  6;         (6)   Ax  +  By  +  C  =  0;         (c)       -     =  1. 


2.  Find  the  equations  of  the  lines  satisfying  the  conditions  given  below 
and  keep  the  results  for  later  use.     Draw  each  line. 

(a)   Through  the  point  (2,  1)  with  a  slope  of  2. 
(6)   Through  the  point  (  —  3,  4)  with  a  slope  of  —  f. 

(c)  Through  the  points  (1,  1)  and  (0,  2). 

(d)  Through  the  points  (-4,  1)  and  (3,  8). 

3.  Find  the  equation  of  the  lino  through  (1,  5)  and  parallel  to  the  line 
hi  (c)  above. 

4.  Find  the  equation  of  the  line  through  (2,  3)  and  at  a  distance  2£ 
from  the  origin.     Draw  the  line  or  lines. 

5.  Find  the  equation  of  the  line  through  (—3,  1)  and  at  a  distance  3 
from  the  origin.     Draw  the  line  or  lines. 

6.  What  is  the  normal  equation  of  the  line  in  (6)  above? 

7.  What  is  the  equation  of  the  line  whose  intercepts  are  a  =  3,  6  =  1? 

8.  Reduce  all  the  equations  obtained  in  2  to  intercept  form. 

9.  What   is  the    angle   between   the   lines,  2x  —  3y  —  1=0   and 
x  -  4  y  =  3? 


162    THE  LINEAR  FUNCTION  AND  THE  STRAIGHT  LINE 

10.  What    is    the    angle    between    the   lines    x/8  —  y/2  =  1    and 
x  cos  30°  +  y  sin  30°  =  4? 

11.  What  is  the  distance  of  (1,  2)  from  each  line  in  Ex.  10? 

12.  What  is  the  normal  equation  of  the  line  whose  intercepts  are 
a  =  -2,  b  =  81 

13.  The  vertices  of  a  triangle  are  (1,  2);    (-3,  4);    (4   7).    Find  the 
perimeter.    Save  all  results. 

14.  What  is  the  altitude  of  each    ertex  of  the  triangle  in  Ex.  13? 

15.  What  is  each  angle  of  the  triangle  in  Ex.  13? 

16.  What  is  the  equation  of  the  line  through  the  origin  and  making  a 
45°  angle  with  the  z-axis? 

17.  What  is  the  equation  of  the  line  through  (3,  5)  and  parallel  to  the 
line  3  x  -  4  y  =  2? 

18.  What  is  the  equation  of  the  line  through  3,  5  and  perpendicular  to 
the  line  in  Ex.  17? 

19.  Find  the  coordinates  of  the  points  which  divide  ths  segment  join- 
ing (—8,  4)  to  (3,  —5)  into  four  equal  parts. 

20.  What  are  the  coordinates  of  the  point  midway  between  (3,  5)    nd 
the  line  y  —  2  cc  +  8  =  0. 

21.  Show  that  the  bisector  of  an  angle  of  a  triangle  cuts  the  opposite 
side  into  segments  having  the  same  ratio  as  the  sides  adjacent  the  bisected 
angle. 

22.  Find  the  equation  of  the  locus  of  all  points  equally  distant  from 
the  points  (2,  3)  and  (4,  1). 

23.  A  square  field  has  a  tree  near  its  center.     Its  distances  from  three 
corners  are  7  rds.,  8  rds.  and  13  rds.,  respectively.    Find  the  side  of  the 
field. 

24.  A  regular  pentagon  has  one  vertex  at  the  origin  and  one  side  in  the 
x-axis.    The  length  of  a  side  is  24.    Find  the  equations  of  the  lines  in 
which  all  the  sides  lie,  respectively. 


CHAPTER  XII 


103.  An  equation  is  said  to  be  explicit  for  y  if  it  is  solved  for 
y  in  terms  of  the  other  variables  and  constants.     If  x  and  y  are 
the  variables  in  the  equation,  it  is  explicit  for  y  if  it  is  of  the 
form 

y  =  f  (*), 

where /(x)  is  any  functio::  of  x.    The  forms 

x2  +  y2  =  25    and    z3  +  3:n/  +  4  =  0 

and  more  generally 

F(*,y)=0, 

are  called  implicit  equations  or  functions  in  x  and  y.  In  such 
cases  y  is  said  to  be  an  implicit  function  of  x  and  x  an  implicit 
function  of  y,  as  is  most  convenient.  By  solving  the  equation 
for  y  or  for  x  it  becomes  explicit. 

104.  Explicit  quadratic  function.  —  y  =  ax2  +  bx  +  c.   The 
graph  of  an  equation  of  this  form  was  constructed  in  34.     An- 
other example  will  now  be  discussed.     Consider 

y  =  X*  -f  2  X  +  3. 

x  =  -4,  -3,  -2,   -1,  -0,  1,  2,    3. 
y  =      5,      0,  -3,  -4,  -3,  0,  5,  12. 

The  graph  is  shown  in  Fig.  83.  Answer  the  following  ques- 
tions: 

1.  Does  the  curve  pass  through  the  origin?    How  is  this 
determined  from  the  equation? 

2.  Determine  the  intercepts  on  both  axes. 

3.  Does  any  branch  of  the  curve  extend  indefinitely  from 
the  origin  ?    That  is,  does  the  curve  have  infinite  branches? 

163 


164 


EQUATIONS  AND  THEIR  GRAPHS 


4.  Is  the  curve  a  closed  curve? 

5.  Is  the  curve  symmetrical  about  either  axis  or  about  any 
other  line  or  about  the  origin  ? 

6.  What  values,  if  any,  of  either  variable  must  be  excluded? 
That  is,  what  values  do  not  correspond  to  points  on  the  curve? 


FIG.  83. 

These  questions  will  now  be  answered  with  reference  to  the 
example  above.  Hereafter  a  discussion  will  include  the  answer- 
ing of  the  above  questions  together  with  pointing  out  any  other 
peculiarities  of  the  curve  and  the  construction  of  the  curve. 

1.  No,  for  x  =  0,  y  =  0  cannot  satisfy  an  equation  having 
a  constant  term. 

2.  At  (0,  —3)  on  the  ?/-axis  and  at  (1,  0)  and  (—3,  0)  on  the 
re-axis. 

3.  Yes.    For   as  x  increases   indefinitely  y  also   increases 
indefinitely. 

4.  No. 

5.  Yes,  about  a  line  parallel  to  the  y-a\is  and  through 
(-1,  0). 


IMPLICIT  QUADRATIC  FUNCTIONS 


165 


6.  No  values  of  x  are  excluded.  All  values  of  y  <  —4  are 
to  be  excluded.  For  such  values  of  y,  x  is  imaginary,  as  can  be 
seen  by  substituting  in  the  equation. 

What  are  the  roots  of  the  equation  z2  +  2  z  —  3  =  0?  Com- 
pare these  values  with  the  it-intercepts. 

1.  Discuss  y  —  xz  —  10  x  +  5. 

2.  " 


105.   Implicit  quadratic  functions.  —  Three  important  cases 
will  be  considered: 

(a)  9  x*  -  16  y2  -  144  =  0. 

Solving  for  y, 

y  =  ±  f  Vz2  -  16. 

x  =-     -8,      -6,      -5,  -4,  -2,      0. 
y  =  ±5.2,  ±3.1,  ±2.2,  ±0,  ±i,  ±i. 
x  =     2,      4,         5,         6,         8. 
y  =  ±i,  ±0,  ±2.2,  ±3.1,  ±5.2. 

Y 


FIG.  84. 

Here  i  is  used  to  denote  that  the  value  is  imaginary.  The 
graph  is  given  in  the  figure.  Let  the  student  discuss  fully. 
This  curve  is  called  a  hyperbola. 

(6)  xy  =  12. 

1,      2,      3, 


x  = 


I) 


4,  6,     12. 

y=      48,      24,      12,      6,      4,      3,  2,      1. 

*  =  -  i  -  i   -  1,   -2,   -3,   -4,  -6,  -12, 

y  =  -48,   -24,  -12,   -6,   -4,   -3,  -2,  -1. 


166 


EQUATIONS  AND  THEIR  GRAPHS 


The  graph  is  given  in  the  figure.     Let  the  student  discuss  fully. 

This  curve  is  also  a  hyperbola: 

(c)  9z2+16?/2  =  144. 

z=-5,   -4,      -3,     -2,   -1,      0,      1,         2,      3,4,    5. 

y=  ±i,  ±0,  ±1.9,  ±2.6,  ±3,  ±3, .±3,  ±2.6,  1.9,  0,  ±i. 

Y 


Y 

FIG.  86. 

The  graph  is  shown  in  the  figure.    Let  the  student  discuss 
fully.     This  curve  is  called  an  ellipse. 

106.  The  curves  of  104,  105,  34,  are  curves  of  the  second 
degree.  They  are  called  conic  sections.  The  student  should 
learn  to  recognize  these  curves  and  their  equations. 


RATIONAL  FRACTIONAL  FUNCTION  167 

107.   Functions  of  the  third  degree  will  be  illustrated  by 

y  =  2  x3  —  5  x2  +  x  +  2. 

1,      f,  2,  3. 
0,  -i,  0,  7. 


y  =  -3,      0,  1,  f, 


FIG.  87. 

* 

The  graph  is  given  in  the  figure.     It  is  typical  of  all  explicit 
cubic  functions.     Let  the  student  discuss  fully. 

108.   Rational  fractional  function.  —  Consider  the  equation 
1  1 


2z~  z  (z  -  1)  (z  -  2) 
The  second  form  is  more  convenient  for  calculating: 

z^-oo,     -2,  -1,     -J,        0,      },  1,        1,      f,        2,3,    4,oo. 

y=  o,  -,v,  -I,  -A,  TOO,  3^-,  f  ±00,  -|,  TOO,  i,  ^T,  o. 
The  graph  is  shown  in  the  figure.  It  should  be  noted  that  the 
infinite  values  of  y  occur  at  the  zero  values  of  the  denominator. 
At  these  values  the  definition  of  continuity  does  not  hold.  The 
function  is  said  to  be  discontinuous  at  such  points.  The  value 
of  x  which  gives  an  infinite  value  of  y  is  called  an  infinity  of  the 


168 


EQUATIONS  AND  THEIR  GRAPHS 


function  or  a  pole  of  the  function.     Let  the  student  discuss  the 
example  fully. 

Y 


X- 


Y' 
FIG.  88. 

109.   As  an  example  of  irrational  functions,  consider 

y  =  Vx?  -  5  x2  +  6  x  =  Vx  (x  —  2)  (as  —  3). 


FIG.  89 


EQUIVALENCE  OF  EQUATIONS  169 

The  second  form  is  most  convenient  for  calculating: 

x  =  Q,        1,  i        2,       I,     3,          4,  5._ 

y  =  0,  ±V2,  ±  V|,  ±0,  ±i,  ±0,  ±  V8,  ±  V30. 

The  curve  is  shown  in  the  figure.     Let  the  student  discuss  fully. 
Construct  the  graphs  and  discuss  fully  the  following: 

1.  y  =  x*-x~1  +  5.  7.   7/2  = 


TC  J 

2.  xy  +  y*  =  23.  8.   xy  +  y2  =  0. 

3.  y  =  x$.  9.   z  -  1  =  —  ! 


4.  y  =  V2  x3  +  x"  -  2  x  +  2.     10.   y  =  (x  -  3)3. 

g          (1      yZ\   y    —    y-    _|_    g  ]_]^       ^    = 

6.  y  = 

[  110.  Simultaneous  equations  of  the  second  and  higher 
degrees  in  two  unknowns  can  be  solved  (at  least  approximately) 
by  means  of  their  graphs.  To  do  this  construct  the  graphs  of 
both  equations  on  the  same  axes  and  to  the  same  scale.  The 
measured  coordinates  of  the  points  of  intersection  of  the  graphs 
will  be  the  solution  of  the  pair  of  equations.  The  slide  rule 
and  tables  of  squares  and  cubes  should  be  employed  to  facilitate 
calculation. 

1.  y  +  xz  =  7  and  y2  +  x  =  11,  find  x  and  y. 

2.  xz  +  y2  =  25  and  x  +  y  +  1  =0,  find  x  and  y. 

3.  xz  +  y-  =  25  and  z2/9  +  2/2/36  =  1,  find  x  and  y. 

4.  z2/9  +  2/Y36  =  1  and  z2/4  -  y*/lQ  =  1,  find  x  and  y. 

5.  y  =  x3  and  xz  +  y2  =  25,  find  x  and  y. 

6.  x2  +  i/2  =  9  and  i/  =  sin  x,  find  Z  and  ?/. 

7.  y  =  cos  x  and  ?/  =  sin  x,  find  z  and  y. 

111.  The  question  of  equivalence  of  equations  and  systems  of 
equations  will  not  be  discussed  systematically.  A  few  examples 
illustrating  the  meaning  of  the  term  and  impressing  the  need  of 
care  in  checking  of  results  will  be  given. 


170  EQUATIONS  AND  THEIR  GRAPHS 

Two  equations  in  the  same  unknown  are  equivalent  when 
every  root  of  each  is  a  root  of  the  other.  Thus 

x2  -  3  x  -  4  =  0    and    5  z2  -  15  x  -  20  =  0 

are  equivalent.  Let  the  student  solve  and  verify  the  state- 
ment. The  equations 

are  not  equivalent.  Let  student  solve  and  verify.  The  follow- 
ing operations  on  an  equation  lead  in  general  to  an  equivalent 
equation. 

1.  Multiplication  or  division  of  both  members  by  the  same 
known  number. 

2.  Addition  or  subtraction  of  the  same  expression  on  both 
sides  of  the  equation. 

3.  Clearing  of  fractions  in  most  ordinary  cases. 

The  following  operations  may  not  lead  to  an  equivalent 
equation. 

4.  Multiplication  or  division  of  both  members  by  an  expres- 
sion containing  the  unknown  (except  as  noted  in  (3)). 

5.  Clearing  an  equation  of  radicals. 

1.  Solve  Vx-\-  5  —  Vx  —  4  =  9;  then  solve  Vx  +  5  + 

Vx  —  4  =  9  and  test  results  by  substitution  in  the  original 
equations. 

2.  Are  x  —  7  —  Vx  —  5  =  0  and  xz  —  15  x  +  54  =  0  equiv- 
alent? 

A  system  of  simultaneous  equations  is  equivalent  to  another 
system  if  every  solution  of  each  system  is  a  solution  of  the  other 
system.  In  solving  systems  of  equations  it  is  best  to  substitute 
all  results  hi  the  original  equations. 

1.  Show,  by  solving  and  substituting,  that  the  system 

x  —  y  =  1      .  $  x  —  y  =  1 

is  equivalent  to        * 


EQUATIONS  171 

2.   Determine  whether  the  systems 


9z-6?/  +  12  =  0 
are  equivalent. 

112.  Some  cases  of  systems  of  quadratic  equations  in  two 
unknowns  are  easily  solved.  A  few  commonly  occurring  cases 
will  be  given  below: 

(a)  One  equation  linear  and  one  quadratic.  This  case  was 
given  in  Chap.  I,  10. 

(6)  Both  equations  of  the  second  degree  and  containing  only 
the  squares  of  the  unknowns.  Thus 

ax2  +  by2  =  c, 
a'x2  +  b'y2  =  c'. 

First,  regard  these  equations  as  linear  in  which  x2,  yzt  instead  of 
x,  y,  are  the  unknowns.  Solve  for  the  values  of  x2,  y2.  Take 
the  square  root  of  each  value  of  x2,  y2  for  the  values  of  x,  y. 

(c)  All  terms  containing  unknowns  are  of  the  second  degree 
but  not  necessarily  the  squares  of  the  unknowns.  Thus 

x2  +  xy  =  10, 
y2  -  xy  =  12. 

Substitute  y  =  mx  in  both  equations  and  get 

x2  +  mx2  =  10, 
m2x2  —  mx2  =  12. 

Solving  each  of  the  last  equations  for  x2  and  equating  results, 

2=      10  12 

m  +  1      m2  —  m 

Solving  the  last  equation  for  m, 

m  =  2.652    and    m  =  -0.452. 
Whence  by  the  equation  y  =  mx,  there  is  obtained 

y  =  2.652z  and  y  =  -0.452  x. 


172  EQUATIONS  AND  THEIR  GRAPHS 

Substituting  the  values  of  m  in  x2  =  — — ;  in  succession, 

m  +  1 

x2  =  2.738 
whence  x  =  ±1.652. 

Substituting  the  value  of  x  in  y  =  2.652  x  gives 

y  =  ±4.381. 

Similarly  using  the  other  value  of  m  will  give  other  values  of 
x,  y.  These  should  be  tested  by  substituting  in  the  original 
equations. 

x2  +  yz  =  25 


(d)   To  solve  the  system  , 

(  oxy  =  66 

Divide  the  second  equation  by  3  and  add  to  the  first  equation, 
obtaining 


Taking  the  square  root 

x  -f-  y  =  ±6  (two  equations). 
By  subtracting  instead  of  adding  as  above  there  is  obtained 

xz  —  2  xy  +  yz  =  14. 
Taking  the  square  root 

x  —  y  =  ±  Vl4  =  ±  3.74+  (two  equations). 

Solving  now  these  four  equations  of  the  first  degree  in  pairs  will 
give  the  desired  solution.  Test  the  results  in  the  original 
equations. 

(e)   The  system 

(rf  +  ff-8 

i  *  +  y  =2 

can  be  solved.  First  divide  the  first  equation  by  the  second, 
member  by  member, 

x*  —  xy  +  ?y2  =  4. 

This  equation  together  with  the  second  equation  form  a  system 
like  the  one  described  hi  (a).  This  system  has,  therefore,  been 
treated  in  Chapter  I.  Complete  the  solution  and  test  results. 


EQUATIONS  173 

i     £2  4-     if  =  25 

1.  Solve  the  system       j  4a;2  + 6?/2  =  ^ 

(  -r2  -J-  ifl  = 

2.  Solve  the  system 

3.  Solve  the  system 

4.  Solve  the  system 

5.  Solve  the  system 

6.  A  piece  of  cloth  on  being  wet  shrinks  10  per  cent  in  length 
and  6  per  cent  in  width.     The  total  loss  of  area  is  5  sq.  yd. 
How  many  square  yards  were  in  the  piece  originally? 

7.  If  $1000  at  a  certain  rate  for  a  certain  time  at  simple 
interest  amounts  to  $1250  (principal  and  interest)  and  if  the 
same  principal  for  3  years  less  time  at  2  per  cent  lower  rate 
amounts  to  $1200,  find  the  rate  and  the  tune. 

8.  A  pole  stands  on  a  tower.     A  man  5'  high  (to  his  eye) 
standing  on  level  ground  finds  that  at  a  certain  distance  from 
the  foot  of  the  tower  the  angle  subtended  (at  his  eye)  by  the 
tower  is  the  same  as  that  subtended  by  the  pole.     The  tower  is 
50'  high  and  the  pole  80'  high.     Find  the  distance  of  the  man 
from  the  foot  of  the  tower. 

9.  A  garden  plot  adjacent  a  wall  is  to  be  fenced.     The  area 
is  to  be  160  sq.  yds.    The  length  is  to  the  breadth  as  3,  2.     Find 
the  dimensions  of  the  plot.     (Fence  on  three  sides.)     Two 
solutions. 

10.  The  sum  of  the  squares  of  two  numbers  is  83.     Their 
difference  is  4.     Find  the  numbers. 

11.  The  area  of  a  circular  race  track  is  150,000  sq.  ft.     The 
inner  diameter  is  to  the  outer  diameter  as  14  to  15.     Find  the 
inner  and  outer  diameters  of  the  track. 

12.  A  rectangle  has  an  area  of  135  sq.  rds.     If  lines  are 
drawn  from  two  opposite  vertices  to  the  diagonal  joining  the 
other  vertices  divide  that  diagonal  into  three  equal  parts.     Find 
the  dimensions  of  the  rectangle. 


CHAPTER  XIII 


P=P' 


TRANSFORMATION   OF  COORDINATES 

113.  It  is  of  great  advantage,  in  certain  problems,  to  be  able 
to  simplify  or  to  change  the  form  of  an  equation.  Certain 
troublesome  terms  may  be  removed  or  some  other  change  may 
be  made.  One  common  way  of  attaining  these  results  is  to 
substitute  for  the  variables  certain  linear  functions  of  new 
variables.  This  is  called  a  linear  transformation.  Such  a 
transformation  has  the  effect  of  moving  the  axes  to  a  new  posi- 
tion with  reference  to  the  curve  whose  equation  is  thus  trans-, 
formed  without  affecting  the  fundamental  properties  of  the 
curve  or  the  degree  of  the  equation. 

(a)  To  rotate  the  axes  of  coor- 
dinates through  an  assigned  angle, 
6,  without  moving  the  origin.  The 
equations  of  transformation  can  be 
determined  by  considering  only  one 
point.  For  all  points  will  be  sim- 
ilarly affected  by  the  transforma- 
tion. Consider  the  point  P(x,  y) 
referred  to  the  axes  OX,  OY.  Let 
the  coordinates  of  the  same  point 
be  x',  y'  referred  to  the  new  axes,  OX',  OY'.  In  the  new  posi- 
tion call  P  =  P'(x',  y'}  which,  of  course,  is  the  same  point  as 
P(x,  y}  referred  to  the  old  axes. 

From  the  figure  x  =  Om,  y  =  mP,  x'  =  01,  y'  =  IP,    Hence 

x  =  Om  =  On  +  nm. 

But  On  =  x'  cos  6    and    nm  =  y'  sin  0. 

Therefore 

(1)  x  =  x'  cos  6  —  y'  sin  6. 

174 


y 
FIG.  90. 


LINEAR  TRANSFORMATION  175 

In  a  similar  manner, 

(2)  y  =  x'  sin  6  +  y'  cos  6. 

Equations  (1)  and  (2)  are  the  ones  desired.  When  6  is  given 
x  and  y  are  expressed  as  linear  functions  of  x'  and  y'.  "When  6 
is  not  known  it  may  be  found  when  certain  other  conditions  are 
given. 

(6)   To  move  the  origin  without  changing  the  direction  of 


the  axes. 

It  is  easily  seen  from  the  figure 
that 


Y' 


FIG.  91. 


(3)  x  =  x'  +  h 

and  

(4)  y  =  y'  -\-  k, 

where  the  new  origin  is  the  point 
O'(h,  k),  referred  to  the  old  axes. 

1.  By  use  of  (3),  (4),  move  the  origin  to  the  point  (2,  3)  in 
the  equation 

xz  +  y2  -  4  x  -  6  y  =  12. 

Substituting  x  =  x'  +  2,  y  =  y'  +  2,  into  this  equation, 

(xr  +  2)2  +  (yf  +  3)2  -  4  (x'  +  2)  -  6  (yr  +  3)  =  12. 
Expanding  and  collecting,  this  reduces  to 

z'2  +  y'2  =  1. 

The  primes  may  now  be  dropped  if  we  remember  that  this 
equation  is  to  be  referred  to  the  new  axes.  Hence  the  last 
equation  may  be  written : 

x2  +  y*  =  1. 

2.  Using  (1),  (2)  rotate  the  axes  45°  in  the  positive  direction 
in  the  equation  xy  =  12. 

Substituting  the  values  of  x,  y  from  (1),  (2),  for  6  =  45°,  into 
this  equation 

(x'  cos  45°  +  y'  sin  45°)  (x'  sin  45°  -  y'  cos  45°)  =  12, 


176  TRANSFORMATION   OF  COORDINATES 

or 

(0.707  x'  +  0.707  y'}  (0.707  x'  -  0.707  y')  =  12, 
or 

0.5z'2-0.5*/'2  =  12, 

or  dropping  primes, 

a?  -  y*  =  24. 

This  equation  represents  the  same  curve  referred  to  the  new 
axes. 

3.  Determine  h,  k  so  that  the  first  power  terms  in  x  and  y 
shall  disappear  from  the  equation 

z2  —  4  1/2  —  2x  -\-  &y  =  1. 

Note.  —  After  substituting  from  equations  (3),  (4),  collect 
the  coefficients  of  the  first  power  of  x  and  equate  the  result  to 
0.  Similarly  equate  the  coefficient  of  y  to  0.  The  resulting 
two  equations  will  determine  h  and  k. 

4.  Determine  6  so  that  the  xy  term  shall  disappear  from 

xz  +  xy  +  2  y*  +  x  =  0. 

Note.  —  Use  Equations  (1)  and  (2)  and  proceed  as  in  the 
last  exercise,  putting  the  coefficient  of  xy  equal  to  zero  and  solv- 
ing for  6. 

5.  In  y  =  mx  +  &,  rotate  the  axes  90°  in  the  positive  direc- 
tion.    What  is  the  meaning  of  m  in  the  new  equation?    Of  6? 

6.  In  ^  +  |T  =  1,  rotate  the  axes  30°.    Note  the  form  of 

AO      y 

the  resulting  equation. 

7.  Using  the  first  of  equations  (3)  determine  h  so  that  the 
term  in  xz  shall  disappear.     Draw  graph  of  original  and  of  new 
equation 


8.  By  any  method  above  remove  the  xy  term  from 

xz  -  f  +  2  xy  +  3  =  0. 

X2          2/2 

9.  Change    TZ  +  q  =  1»   *°    polar   coordinates,   using  the 

transformation  x  =  r  cos  0,  y  =  r  sin  6  (73,  Ex.  1J,  12). 

10.  Change  y*  =  8  x  to  polar  coordinates. 


CHAPTER  XIV 
CONIC   SECTIONS 

114.  The  conic  sections  constitute  the  geometric  aspect  of 
integral  functions  of  the  second  degree.     In  these  are  included 
all  integral  equations  in  two  variables  where  at  least  one  term 
is  of  the  second  degree.     For  this  reason  the  conic  sections  are 
called  loci  of  the  second  order,  curves  of  the  second  order,  or 
curves  of  the  second  degree. 

Historically  the  geometric  idea  was  developed  first.  The 
correlation  of  the  geometric  with  the  algebraic  (analytic)  idea 
with  respect  to  these  curves  has  been  of  much  value  in  the  study 
of  the  laws  of  nature. 

Definition.  —  The  locus  of  a  point  which  moves  in  a  plane 
so  that  the  ratio  of  its  distances  from  a  fixed  point  and  a  fixed 
straight  line  is  constant  is  called  a  conic  section  *  or  for  short  a 
conic. 

This  definition,  based  on  the  discovery  of  the  property  of 
these  curves  by  the  Greeks,  furnishes  a  convenient  starting 
point  for  an  introductory  analytic  study  of  the  curves. 

The  fixed  point  referred  to  in  the  definition  is  called  the  focus 
of  the  conic,  and  the  fixed  line  the  directrix  of  the  conic. 

The  conies  are  conveniently  classified,  for  purposes  of  ele- 
mentary study,  according  to  the  different  values  which  the  ratio, 
referred  to  in  the  definition,  may  take.  This  ratio  is  called 
the  eccentricity  of  the  conic  and  will  be  denoted  by  e. 

115.  Ratio  equal  to  one,  e  =  1.    Parabola.  —  Let  F  be  the 
focus,  DD'  the  directrix  and  P  (x,  y)  any  point  on  the  curve. 

*  For  proof  that  the  curves  of  intersection  of  planes  with  a  right  circular 
cone  have  this  property,  see  Wentworth  Geom.,  Rev.  Ed.,  p.  458,  or  some 
treatise  on  conic  sections. 

177 


178 


CONIC  SECTIONS 


We  wish  to  find  the  equation  of  this  curve.     Choose  the  origin 
on  the  curve  midway  between  the  focus  and  the  directrix. 

From    the    hypothesis    and   the 
figure  we  can  write 

FP  =  KP,  OF  =  MO  =  p. 
Expressed  in  terms  of  x,  y  and  p 
the  relation 

FP  =  KP 
becomes 


FIG.  92. 


or  (1)  ?/2  =  4  px. 

This  is  the  desired  equation  of  the 
parabola  in  the  standard  form.  Let 
the  student  discuss  the  curve.  This  is  the  same  kind  of  curve 
as  the  one  given  in  104,  the  difference  being  the  position  with 
reference  to  the  axes. 

1.  Find  the  total  distance  across  the  curve  through  the  focus 
perpendicular  to  the  aj-axis.     This  double  ordinate  through  the 
focus  is  called  the  latus  rectum. 

2.  Move  the  origin  to  the  point  (h,  k)  and  note  the  change 
in  the  equation. 

3.  Rotate  the  axes  90°  in  the  negative  direction  and  note  the 
form  of  the  resulting  equation. 

4.  Find  the  equation  of  a  parabola  (standard  form)  that 
passes  through  (5,  2).     Determine  the  distance  from  the  focus 
to  the  directrix  and  the  distance  from  the  focus  to  the  origin-. 

5.  Transform  equation  (1)  so  the  origin  will  be  at  F. 

6.  Transform  the  resulting  equation  of  example  5,  to  polar 
coordinates  with  pole  at  focus  and  polar  axis  the  z-axis.     Note 
form  of  equation. 

7.  Show  that  for  any  two  points  on  a  parabola,  the  squares 
of  the  ordinates  are  proportional  to  the  abscissas  (equation  to 
be  in  standard  form).     Give  a  geometric  interpretation  of  this 
theorem,  that  will  be  independent  of  the  position  of  the  axes  or 
the  form  of  the  equation  used. 


ELLIPSE 


179 


8.  Simplify  the  equation  y  =  4  z2  —  6  x  +  40,  by  putting  it 
in  the  form  y  =  kx2. 

Note.  —  The  constant  and  the  term  in  x  may  be  removed  by 
use  of  equation  (3),  (4),  113. 

9.  Simplify  and  discuss  4  ?/2  —  6  £  +  3  =  x  —  5. 

10.  Derive  the  equation  of  a  parabola  passing  through  (1,  1), 
(3,  5),  (0,  0).     Assume  equation  of  form  (y  —  a)2  =  k  (x  —  I) 
and  determine  a,  k,  I  and  write  the  equation  accordingly. 

116.  Ratio  less  than  one,  e  <  1.  Ellipse.  —  Use  the  nota- 
tion of  115  except  that  MF  =  p  =  MA  +  AF,  MA  ^  AF. 
Now  from  the  figure  and  the  hypothesis  we  can  write 


AF 


-      _    ^1 

~         ~ 


MA  ~  KP 
AF  =  e-MA. 


Solving  for  MA,  AF,    MA  = 


AF  = 


1+6 

ep 

r+v 


Now  FP  =  e  •  KP  for  all  points  of  the  curve.     Expressing  this 
relation  in  terms  of  the  coordinates  of  P  (x,  y)  there  is  obtained 


180  CONIC  SECTIONS 

which  reduces  by  clearing  of  radicals  to 

(1)  (1  -  e2)  z2  -  2  epx  +  yz  =  0. 

This  is  the  equation  of  the  ellipse  in  the  position  shown  in  the 
figure. 

1.  Discuss  the  curve  from  the  above  equation. 

2.  Move  the  origin  to  the  point  [•_   a,  0  j  to  remove  the 
term  in  the  first  power  of  x.     (See  113  (6).) 

CT) 

3.  Call  1  —  —  g  =  a  and  (1  —  e2)  a2  =  b2  in  the  equation  ob- 

J.          €> 

tained  in  example  (2)  and  show  that  the  equation  reduces  to 

(2)  -2  +  ^=1- 

a2      o2 

This  is  the  standard  form  of  the  equation  of  the  ellipse. 
The  origin  is  now  0'  and  the  y-axis  is  O'Y'  in  the  figure  above. 

4.  Note  the  meaning  of  a,  b  in  the  figure  and  show  that  the 
intercepts  on  the  x-axis  are  —a  and  +a,  and  on  the  ?/-axis  the 
intercepts  are  —6  and  +6.    The  values  a  and  6  are  the  semi- 
axes  of  the  ellipse.     The  value  a  is  the  semimajor  axis  and  the 
value  6  is  the  semiminor  axis;  0'  is  the  center. 

5.  Discuss  the  curve  by  use  of  the  equation  (2)  above. 

a  * 

6- 


Solve  this  equation  for  p  in  terms  of  a  and  e.    The  result  is 

a  (1  -  e2) 

p  =  —i  -  '-  =  MF. 
e 

7.  Find  the  value  of  MA  in  terms  of  a  and  e. 

8.  If  c  —  ae,  show  by  use  of  the  values  of  a  and  6  above  that 
a2  —  62  =  c2  for  any  ellipse. 

9.  Show  by  use  of  the  result  of  Ex.  (8)  that  if  e  =  0  the 
ellipse  becomes  a  circle. 

Note.  —  This  supposition  makes  a  =  b. 

10.  Find  the  equation  of  the  locus  of  a  point  that  is  always 
16  units  from  the  point  (3,  1). 


HYPERBOLA  181 

11.  A  circle  has  its  center  at  (3,  2)  and  passes  through  the 
point  (8,  11).     Find  its  equation. 

Note.  —  Formulate  the  distance  between  (3,  2)  and  any  point 
(x,  y)  of  the  curve  and  equate  this  expression  to  the  given 
radius.  Free  the  equation  of  radicals. 

12.  Find  the  equation  of  the  ellipse,  in  standard  form,  if  the 
eccentricity,  e,  is  §  and  the  curve  passes  through  (3,  4). 

Note.  —  Form  two  equations  and  determine  a  and  b  of  equa- 
tion (2). 

13.  Find  a,  b,  c,  e  and  p  in  the  ellipse  25  x*  +  144  y2  =  1500. 

14.  Prove  that  for  any  ellipse  FP  +  F'P  =  2  a  where  P  is 
any  point  on  the  curve. 

15.  What  is  the  standard  equation  of 


/v»2  ^y2 

16.  With  the  standard  equation  -^  +  j-  =  1,  move  the  origin 

y      ^i 

to  the  left-hand  focus  (—  c,  0),  then  change  to  polar  coordinates. 
Note  the  form.     Compare  with  Ex.  (6),  115. 

17.  Find  the  length  of  the  double  ordinate  through  one  focus 

x2      7/2 

in  -5  +  r;  =  1,  and  find  the  distance  of  the  end  of  this  ordinate 
a2      o2 

from  0,  and  from  the  other  focus.     See  Ex.  1,  115. 

117.  Ratio  greater  than  one,  e  >  1.  Hyperbola.  —  With 
the  same  notation  as  in  the  last  section  and  from  the  annexed 
figure  may  be  written 

(1)  (e2  -  1)  z2  +  2  epx  -  y*  =  0,     e  >  1, 

where  the  origin  is  at  A  and  where  M  A  +  AF  =  p. 

Since  (1  —  e2)  <  0,  move  the  origin  to  (    _   „,  01.    The  above 

equation  becomes 


or  a?  - 

1  -  e2      (e2  -  I)2 


182 


CONIC  SECTIONS 


Calling  ez  _  i  =  a>  and  a2  (e2  —  1)  =  62,  the  last  equation  may 
be  put  hi  the  form 

A  A 

(2) 


.K_. 


Mj  0 

-p— > 

— e- 


Da 


a-->*-a- 


» 


D 
FIG.  94. 


This  is  the  standard  form  of  the  equation  of  the  hyperbola 
referred  to  0'  is  the  origin  and  O'Y'  the  y-axis. 

1.  If  c  =  ae,  show  that  a2  +  fr2  =  c2. 

2.  Discuss  the  curve  from  equation  (2). 

/y»2  qj2 

3.  Discuss  the  curve  whose  equation  is  -5  —  ^  =  —  1,  and 

a2      bz 

compare  with  the  curve  of  equation  (2). 

The  curve  of  this  exercise  is  called  the  conjugate  hyperbola 
of  the  curve  of  equation  (2).  Note  the  intercepts  of  both 
curves  on  the  axes. 


x       y  x       if 

4.   Discuss  the  two  curves  ^  —  77  =  1  and  T^  —  ?r  = 

lo      9  lo       9 


1. 


Draw  the  curves  on  the  same  axes. 

5.  Prove  that  for  any  hyperbola  F'P  —  FP  =  2  a,  equation  (2) . 

6.  With  equation  (2)  rotate  the  axes  45°  in  the  negative 
direction.     Compare  the  result  with  the  example  of  105  (6). 

x2      ii2 

7.  With  —  —  ZT  =  1,  move  the  origin  to  (c,  0)  and  change  to 


DIAMETER 


183 


polar  coordinates.  Note  the  form  and  compare  with  the  polar 
forms  of  the  parabola  and  the  ellipse  previously  derived. 

4)  ft 

/£^  ni£, 

8.  Find  e,  a,  b,  c,  p  for  the  curve  ^=  —  ^  =  1. 

AD      ID 

9.  Find  the  equation,  in  standard  form,  of  a  hyperbola  of 
eccentricity,   e  =  2,   and   passing  through   (7,   5).     Find  the 
equation  of  the  conjugate  hyperbola. 

10.  Find  the  equation  of  a  hyperbola  in   standard  form 
which  passes  through  (3,  5)  and  (5,  7). 

118.  Diameter.  —  A  line  which  bisects  a  system  of  parallel 
chords  of  any  curve  is  a  diameter  of  the  curve. 

All  the  conies  have  diameters.  To  illustrate,  consider  the 
ellipse  whose  equation  is 

^j.^!_i 
a2  '  62 

Let  P'  (x',  y')  be  the  midpoint  of  any  chord,  AB.  Let  P  (x,  y) 
be  a  variable  point  on  AB.  Call  P'P  =  r,  and  consider  r 
positive  if  P  is  between  P'  and  A,  negative  when  P  is  between 
P'  and  B.  Now  from  the  figure  write 


=  sin0. 


FIG.  95. 

If  P  moves  to  A  or  to  B  its  coordinates  must  satisfy  the  equation 
of  the  ellipse.     Solving  the  last  two  equations  for  x}  y  and 

y?      ip 
substituting  in  —z  +  -^  =  1  gives 

€L          0 


2rx/cos0 


?/2  +  2  ry'  sin  e  +  r2  sin2  0  _ 


184  CONIC  SECTIONS 

When  P  is  on  the  curve  the  roots  of  the  last  equation  with 
regard  to  r  are  equal  but  of  opposite  signs.  Hence  the  coeffi- 
cient of  the  first  power  of  r  must  be  0.*  Therefore, 

x'  cos  &      y'  sin  6  _  _ 

~*~       ~~V~ 

— 62 
or  y'  =  — j~  x'  cot  6. 

This  is  a  linear  relation  between  xf,  y',  the  coordinates  of  the 
midpoint  of  any  (consequently  all)  chords  making  'a  given 
angle,  6,  with  the  axis.  The  equation,  therefore,  holds  for  all 
points  of  the  bisector  of  a  parallel  system  of  chords,  and  is  the 
equation  of  the  diameter  which  bisects  these  chords,  that  is 

of  CD. 

xz      yz 

1.  Find  the  equation  of  the  diameter  of  an  ellipse  =^7  +  %  =  1, 

ID      y 

which  bisects  the  system  of  chords  which  make  a  30°  angle  with 
the  re-axis. 

Note.  —  Adapt  the  method  above  to  this  case  and  work  out 
in  full. 

2.  Find  by  a  method  similar  to  the  above,  the  equation  of 
the  diameter  of  the  parabola  which  bisects  all  chords  which 
make  a  45°  angle  with  the  z-axis. 

3.  Prove  that  if  the  slope  of  a  diameter  of  the  ellipse, 

£  +  £-i 

a2  +  6-  ~  L| 

is  m  and  the  slope  of  the  corresponding  system  of  chords  is  m', 

62 

then  m,'  m'  =  -=• 
a2 

4.  Show  that  if  the  diameter  of  an  ellipse  has  a  slope  m,  and 
the  slope  of  the  bisected  chords  is  m',  then  the  diameter  which 
bisects  the  chords  parallel  to  the  first  diameter  has  m'  for  its 
slope. 

The  two  diameters  referred  to  in  this  exercise  are  called 
conjugate  diameters. 

*  See  92,  Ex.  5. 


ECCENTRIC  ANGLES 


185 


x       iiz 

5.  Find  the  equation  of  diameter  of  -»  +  ^  =  1,  which  passes 

y      xo 

through  the  point  (1,  1)  and  find  the  equation  of  the  diameter 

conjugate  to  this. 

xz      y-i 

6.  Find  the  equation  of  the  diameter  of  ^  —  ^  =  1,  which 

bisects  the  chords  which  are  parallel  to  2  x  +  3  y  =  0. 

119.  Eccentric  angles,  (a)  Ellipse.  —  In  the  figure  the  large 
circle  is  called  the  major  auxiliary  circle,  the  small  circle  the  minor 
auxiliary  circle  of  the  ellipse.  P',  P  are  points  on  the  ellipse  and 
major  circle  having  the  same  abscissa.  The  angle,  6  =  A  OP', 
is  called  the  eccentric  angle  of  the  point,  P,  of  the  ellipse. 


FIG.  96. 
From  -z  +  f^  =  1  and  x'2  +  y'2  =  a2,  it  is  evident  that  for 

\Jv  t/ 

x  =  x'  the  corresponding  values  of  the  t/'s  are  related  by 


x       y 
1.  Suppose  the  ellipse  -5  +  r^  =  1  be  cut  into  narrow  strips 

CL          0 

parallel  to  the  y-axis  (approximately  rectangles).     From  the 

x*      y* 
relation  above  show  that  the  area  of  the  ellipse  -|  +  p  =  1  is 


186 


CONIC  SECTIONS 


b/a  times  the  area  of  the  circle  x2  +  y*  =  a2.  That  is,  that  the 
area  of  the  ellipse  is  nab. 

2.  Show  that  the   area   of   the 
ellipse  is  Vl  —  e2  times  the  area  of 
the  circle  given  in  example  (1). 

3.  Show  that  if  <j>  is  the  angle 
between   the   plane   of   the    circle 
z2  +  yz  =  a2  and  the  plane  of  the 

XZ        11" 

ellipse  -5  +  rj  =  1,  then  sin  <j>  =  e 

FIG.  97.  and  cos  <j>  =  Vl  —  e2  and  that  con- 

sequently from  the  results  above  cos  <j>  =  b/a  =  \/l  —  e2. 
(Remember  e  =  c/a,  c2  =  a2  —  &2.) 

4.  Find  the  equation  of  the  ellipse  which  is  the  projection  of 
the  circle  xz  +  yz  =  25  on  a  plane  at  an  angle  30°  with  the  plane 
of  the  circle. 

(6)   Hyperbola. —  In   the   figure   P,  P'  are   corresponding 

x2      y* 
points  on  the  hyperbola  -5  —  jj^  =  1  and  its  auxiliary   circle 

£2  +  y*  =  a2-    The  angle  0  is  called  the  eccentric  angle  of  the 
point  P,  on  the  hyperbola.     It  is  easily  seen  that 
x  =  a  sec  0, 

Y 


FIG.  98. 

These  relations  are  closely  connected  with  a  very  interesting 
set  of  functions,  known  as  the  hyperbolic  functions.    These 


GENERAL  EQUATION  187 

functions  have  properties  quite  similar  to  those  treated  in 
Chapter  VIII.     (See  McMahon  Hyperbolic  Functions.) 

120.   The  most  general  equation  of  the  second  degree  hi  two 
variables  is  of  the  form 

(1)  Ax*  +  Bxy  +  Cy2  +  Dx  +  Ey  +  F  =  0. 

Under  certain  conditions  the  terms  of  the  first  degree  may  be 
removed  by  moving  the  origin.     Assume 

x  =  x'  +  h    and    y  =  y'  +  k 
and  substitute  in  (1)  ;  the  result  is 

(2)  Axf  +  Bx'y'  +  Cy'*  +  (2  Ah  +  Bk  +  Z>)  x' 


+  Ek  +  F  =  0. 

The  terms  of  the  first  degree  can  be  removed  if  and  only  if 
simultaneously  the  equations, 

2Ah  +  Bk  +  D  =  0, 

Bh  +  2  Ck  +  E  =  0, 
hold  for  finite  values  of  h  and  k.     Solving  for  h  and  k  gives 

2CD-BD          ,  2AE  -BD 

~~  ~~ 


These  values  are  finite  if  B2  —  4  AC  =  A  ^  0.     Therefore,  the 

first  degree  terms  can  be  removed  by  moving  the  origin  if 

A  7*  0.    The  term  in  xy  may  always  be  removed  by  rotating 

the  axes.     It  will  then  be  sufficient  to  discuss  in  detail  the 

equation, 

(3)  Ax2  +  Cyz  +  Dx  +  Ey  +  F  =  0. 

I.   Consider  this  equation  when  A  ^  0,  C  ^  0.    Now  A  5^  0. 

The  terms  of  first  degree  can  be  removed.     The  resulting  equa- 
tion will  be  of  the  form 

Ax2  +  Ci/2  +  F'  =  0. 

(a)  If  A  and  C  are  like  signed  and  F'  7*  0,  the  equation 
represents  an  ellipse,  real  if  the  sign  of  F'  is  opposite  that  of  A 
and  C,  imaginary  if  the  sign  of  F'  is  the  same  as  the  sign  of  A 
and  C.  If  F'  is  zero,  the  ellipse  is  a  point  ellipse. 


188  CONIC  SECTIONS 

(6)  If  A  and  C  are  unlike  signed  and  F'  ^  0  the  equation 
represents  a  hyperbola. 

If  F'  =  0,  the  left  member  of  (3)  breaks  up  into  two  lineal* 
factors  and  represents  two  straight  lines  through  the  origin. 

II.  Consider  the  case  where  either  A  or  C  vanishes,  say 
A  =  0  and  C  ^  0.  Now  equation  (2)  becomes 

(4)  Cy  +  Dx  +  Ey  +  F  =  0. 
If  C  =  0  and  A  ^  0,  the  equation  reduces  to 

(5)  Ax*  +  Dx  +  Ey  +  F  =  0. 

By  moving  the  origin  to  a  point  (h,  0)  equation  (4)  may  be 
reduced  to 

(6)  Cyz  +  Dx  =  0, 

and  by  moving  the  origin  to  (0,  k)  equation  (5)  may  be  re- 
duced to 

(7)  Ax*  +  Ey  =  0. 

Equations  (6),  (7)  represent  parabolas. 

In  case  the  equations  (4),  (5)  take  the  form 

(8)  Cy2  +  F'  =  0, 

(9)  Axz  +  F  =  0, 

respectively,  they  represent  pairs  of  straight  lines  parallel  to 
the  axes. 

121.  Confocal  conies.  —  Two  conies  are  confocal  when  they 
have  the  same  foci.  Consider  the  two  conies  whose  equations 
are: 


The  foci  of  these  conies  will  coincide  if  o2  —  62  =  a\2  —  2>i2,  or 
a*  -  ai2  =  62  -  &!2.  Let  ai2  =  o2  +  X  and  6i2  =  62  +  X.  Sub- 
stituting in  (2)  : 

2  ° 


CENTERS  OF  CONICS  189 

The  curve  (3)  is  confocal  with  the  curve  (1)  for  all  values  of  X. 
Why? 

/*»2  nj'Z 

1.  Find  the  equation  of  a  conic  confocal  with  ^  +  ^  =  1 

AO          \j 

and  passing  through  (5,  6). 

xz  y2 

Note.  —  Write  ;     7—  r  +  n  ,   .  =  1.    Substitute  the  coordi- 

ib  +  A    y  +  x 

nates  of  the  given  point  hi  this  equation  and  solve  for  X. 
Having  found  X,  substitute  its  value  in  the  above  equation. 

2.  For  what  values  of  X  will  the  equation 


16  +  X      9  +  X 

represent  an  ellipse?  For  what  values  of  X  will  it  represent 
a  hyperbola?  See  116-117  and  determine  what  values  of  X 
make  this  equation  like  those  referred  to. 

x2      y2 

3.  Find  the  equation  of  a  conic  confocal  with  —  —  ^  =  1 

*7  J.D 

and  passing  through  (5,  7). 

4.  Determine  whether  8  x2  +  12  y2  =  96  and  3  x2  -  8  y2  =  24 
are  confocal  conies. 

122.  Centers  of  conies.  —  The  ellipse  and  hyperbola  haver 
two  perpendicular  axes  of  symmetry  and  therefore  have  a 
center  of  symmetry.  This  point  is  called  the  center  of  the 
conic.  The  circle  is  considered  as  a  special  case  of  the  ellipse. 
When  the  equations  of  the  ellipse  and  hyperbola  are  in  standard 
form  the  centers  are  at  the  origin.  When  the  equations  are  not 
in  standard  form  the  center,  referred  to  the  original  axes,  is 
found  by  the  values  of  h,  k  by  which  the  terms  of  first  degree 
are  removed.  The  values  of  h  and  k  being  the  coordinates  of 
the  center  referred  to  the  original  coordinate  axes.  It  is  to  be 
understood  that  the  xy  term  has  first  been  removed  from  the 
equation  before  applying  this  method. 

Full  treatment  of  the  method  of  finding  the  centers  of  conies 
cannot  be  entered  into  in  this  course.  Some  examples  will  be 
of  use  in  showing  how  to  determine  the  center  of  a  conic  in 
certain  cases. 


190  come  SECTIONS 

(a)   Find  the  center  of  the  circle  x2  +  y*  —  Qx  +  8y  =  0. 
This  equation  may  be  written  in  the  form 

x2  -  6  x  +  9  +  y2-  +  8  y  +  1 6  =  25, 
or  (x  -  3)2  +  (y  +  4)2  =  25. 

It  is  evident  that  if  the  origin  be  moved  to  the  point  (3,  —4) 
that  the  first  degree  terms  will  disappear  and  the  equation 
would  be  in  standard  form.     Therefore,  the  center  of  the  circle 
is  the  point  (3,  —4),  referred  to  original  axes  of  coordinates. 
Let  the  student  draw  the  curve  and  verify  this  result. 
(6)   Find  the  center  of  4  x2  +  6  y2  +  12  x  -  24  y  =  3. 
This  equation  can  be  written  in  the  form 

4x2  +  I2x  +  9  +  6?/2  -  24 ?/  +  24  =  36, 
or  (2x  +  3)2  +  6  (y  -  2)2  =  36. 

It  is  evident  if  the  origin  be  moved  to  (  — f,  2)  the  first  power 
terms  will  vanish  and  the  center  will  be  at  that  point. 

1.  Solve  each  of  the  above  problems  by  the  method  of  mov- 
ing the  origin,  113  (6). 

2.  Show  that  center  of  the  circle  xz  +  yz  +  Dx  +  Ey  +  F  =  0 

is  at  the  point  ("o— i  ~"S~~)' 

3.  Show  that  the  equation  of  the  circle  whose  center  is  (a,  6) 
and  radius  r  is  (x  —  a)2  +  (y  —  6)2  =  r2. 

4.  Find  the  center  of  x2  -  y2  -  3  y  +  24  =  0. 

5.  Find  the  center  of  xz  +  y2  -  Sx  +  IQy  -  20  =  0. 

MISCELLANEOUS  PROBLEMS 

1.  Reduce  2  x*  -  5  xy  -  3  yz  +  9  x  -  13  y  +  10  =  0  to  one  of  the 
standard  forms  and  determine  a,  b,  c,  e,  p  as  the  case  may  require. 

2.  What  is  the  equation  of  the  circle  of  radius  10  and  center  at  the 
point  (2,  3).     (See  Ex.  3,  122.) 

3.  What  is  the  length  of  the  common  chords  of  x2  +  j/2  =  8  and  9  x2  + 
4  y*  =  36?     (The  common  chord  joins  two  points  of  intersection.) 

4.  Find  the  equations  of  the  straight  lines  which  coincide  with  the 
common  chords  in  Ex.  3. 

5.  Find  the  standard  equation  of  the  hyperbola  whose  foci  are  the 
points  (—3,  0)  and  (3,  0)  and  eccentricity  1.5. 


MISCELLANEOUS  PROBLEMS 


191 


6.  See  if  the  curve  ^  =  2  x  —  VT/Q  touches  the  curve  3  z2  —  6  if  =  1. 
Note.  —  Solve  simultaneously  and  determine  whether  the  curves  cut 

or  just  touch  each  other. 

7.  Find  the  equation  in  standard  form  of  an  ellipse  whose  foci  are 
(  —  3,  0)  and  (3,  0)  and  eccentricity  2/3. 

8.  Find  the  equation  of  the  run  of  a  6"  stove  pipe  cut  at  an  angle  of 
30°  with  the  axis  of  the  pipe.     (Standard  form.) 

9.  Find  the  equation  of  the  boundary  of  the  shadow  of  a  circle  of 
radius,  r,  on  a  plane  making  a  45°  angle  with  the  plane  of  the  circle. 
(Standard  form.)     (Light  vertically  above  plane.) 

10.  Find  the  equation  of  a  circ  e  passing  through 
the  points  (1,  2),  (3,  5)  and  (-1,  4)      Find  the 
lengths  of  the  three  chords  joining  these  points. 

Note.  —  Use  (x  —  a)2  +  (y  —  6)2  =  r2  and  deter- 
mine a,  6  and  r. 

11.  Find  the  equation  of  a  parabola,  in  standard 
form,  which  has  its  focus  at  (4,  0). 

12.  If  x  =  vt  and  y  =  —  aP,  eliminate  t  and 
discuss  the  locus  of  the  resulting  equation. 

13.  Prove  that  two  elliptical  gear  wheels  of  the 
same  size  and  shape  will  work  together  smoothly  if 
connected  by  rods  joining  their  foci  as  shown  in  the 
figure. 

14.  Determine  the  kind  of  conic  represented  by 
the  following: 

(a)   4x2  +  yz  -  13z  +  7y  -  1  =  0. 

(6)   3z2  -4j/2  -  6^  +  9  =  0. 

(c)   7  z2  -  17  xy  +  6  y2  +  23  x  -  2  y  -  20  =  0. 

15.  Find  a,  6,  c,  e,  p  and  the  coordinates  of  the  center  of: 

81  =0. 


(a) 

(6)   5z2 

(c)   5x2 


5xy  -  7y2  -  165x 


=  0. 
1320  =  0. 


16.  Find  the  equation  of  a  conic  confocal  with  Tc  +  q  =  1  ^d  passing 
through  the  point  (a),  (6,  5);   (6),  (3,  2). 

17.  In  the  figure  AB  and  CD  are  the  axes  of  the  ellipse.     RT  =  %  AB, 
RS  =  %  CD.    Show  that  if  T  be  made  to  move  on  CD  while  S  moves  on 
AB,  then  R  traces  the  ellipse.     (This  is  the  principle  of  the  ellipsograph.) 

18.  By  use  of  the  property  of  the  curve  expressed  in  Ex.  14,  116, 
devise  a  method  of  constructing  the  curve  by  the  intersections  of  pairs  of 
arcs  whose  centers  are  at  the  foci. 

19.  By  use  of  the  definition  of  the  parabola,  116,  show  how  to  construct 


192 


CONIC  SECTIONS 


a  parabola  by  intersection  of  arcs  whose  centers  are  at  the  focus  with 
straight  lines  parallel  to  the  directrix. 


20.  By  use  of  the  property  of  the  curve  expressed  in  Ex.  5,  117,  devise 
a  method  of  constructing  points  on  a  hyperbola  by  intersections  of  arcs 
whose  centers  are  at  the  foci. 

21.  Show  that  an  ellipse  can  be  drawn  by  using  a  string  of  length  2  a 
fastening  the  ends  of  the  string  at  the  foci  and  holding  a  pencil  against  the 
string  while  drawing  the  curve.     See  figure  below. 


JL--  Pencil  point 


Ex.  21. 


are  the 


22.  Show  that  the  points  (2,  2),  (-2,  -2)  and  (2  V3,  -2 
vertices  of  an  equilateral  triangle. 

23.  Prove  that  (10,  0),  (5,  5),  (5,  -5),  (-5,  5)  are  the  vertices  of  a 
rapezoid. 

24.  In  any  triangle  show  that  a  line  joining  the  middle  points  of  two 
rides  is  parallel  to  the  third  side  and  equal  to  half  of  it. 

Note.  —  Use  two-point  form  and  compare  slopes. 

25.  Find  the  equation  of  a  circle  whose  center  is  at  (3,  2)  and  its  radius  4. 
Discuss  the  curve  y  =  x2  —  2  x  —  3. 

Find  the  point  of  intersection  of  x  —  7  y  =  25  with  xz  +  y2  =  25. 


26. 

27. 
Plot. 

28. 

29. 
tf- 


Change  x2  -f  yz  —  25  to  polar  coordinates. 

Move  the  origin  so  as  to  remove  the  first  degree  terms  from  a?  — 
1  —  12  =  0  and  discuss  the  curve  with  reference  to  the  new  axes. 


CHAPTER  XV 

THEOREMS  ON  LIMITS,  DERIVATIVES  AND    THEIR 
APPLICATIONS 

123.  Theorem  I.*  —  The  limit  of  the  sum  of  two  or  more  f 
infinitesimals,  37,  is  zero. 

Let   81,    52,    53  be  three  infinitesimals.    By  definition  each 
must  become  and  remain  less  than  some  arbitrarily  small 

number,  |  say,  where  e  is  arbitrary.     Hence  at  some  stage  and 
o 

thereafter, 

e  e  € 

8t<§,     82<3,     83  <  3 

and 

81  ~\~  82  -\~  83  <C  3  •  ^  =  e. 

Since  c  is  arbitrary  the  sum  5i  +  52  +  53  satisfies  the  definition 
of  a  limit  and  we  conclude  (see  37) 

Lim  (Si  +  52  +  83)  =  0. 

124.  Theorem  H.  —  The  limit  of  the  sum  of  two  or  more 
variables  is  the  sum  of  their  limits. 

Let  x,  y,  z  be  three  variables  and  a,  6,  c  their  respective 
limits.     Let 

x  —  a  =  Si,     y  —  b  =  52,     z  —  c  =  53.f 

Now  from  the  definition  of  limit,  Si,  S2,  83,  each  has  zero  for  its 
limit  (see  37d).    Write 

x  =  a  +  Si,     y  =  b  +  52,     z  =  c  +  52 

*  This  and  the  three  following  theorems  may  be  treated  as  assumptions 
if  preferred  and  proofs  omitted. 

t  In  this  and  following  theorems  the  term  "  more  "  is  not  to  be  interpreted 
to  mean  an  infinite  number. 

J  If  2  <  c,  83  will  be  negative  and  similarly  for  the  others. 

193 


194  LIMITS  AND  DERIVATIVES 

and  add 

x  +  y  +  z  =  a  +  6  +  c  +  61  +  82  +  53. 

Ihe  limit  of  the  right  side  of  this  equation  is  a  +  6  +  c  (Theo- 
rem I).  Therefore, 

Lira  (x  +  y  +  z)  =  lim  x  +  lim  T/  +  lim  z. 

125.  Theorem  HI.  —  The  limit  of  the  product  of  two  or 

more  variables  is  the  product  of  their  limits. 
With  the  same  notation  as  in  124,  write 

Lim  xyz  =  Lim  (a  +  61)  (6  +  &)  (c  +  53) 

=  Lim  (a&c  +  a8z83  +  b8t83  +  cdi83  +  ab8s  +0,082 
+  bcSj  +  818283)  =  abc, 

since  every  term  on  the  right  after  the  first  has  zero  for  its  limit. 
In  a  similar  way  the  proof  can  be  extended  to  any  number  of 
variables.  The  theorem  is,  therefore,  true. 

126.  Theorem   IV.  —  The  limit  of   the    quotient   of   one 
variable  divided  by  another  is  the  quotient  of  their  respective 
limits. 

With  the  same  notation  as  above  write 

r.     x      T .     a  —  81 

Lim  -  =  Lim  r -• 

y  b  -  82 

By  long  division  the  right  member  may  be  written  so  that 


since  the  numerator  of  the  second  fraction  has  zero  for  its  limit. 
Therefore  by  42,  III,  the  fraction  has  zero  for  its  limit. 

127.  Definition  and  formation  of  the  derivative  of  a  func- 
tion. —  All  numbers  may  be  thought  of  as  values  which  a 
variable  may  assume.  Any  number  may  be  considered  as  the 
sum  of  two  numbers,  one  a  value  which  a  variable  may  assume 
at  some  instant,  the  other  an  increment  to  the  first  so  that  the 
sum  is  a  subsequent  value  of  the  variable.  This  way  of  think- 
mg  of  numbers  and  in  particular  of  symbols  or  variables  which 


DERIVATIVE  195 

assume  number  values  yields  a  very  useful  instrument  for 
solving  certain  kinds  of  problems  which  we  have  hitherto  been 
unable  to  attack. 

In  order  to  make  use  of  this  idea  it  is  necessary  to  learn  to 
calculate  and  to  manipulate  a  function  called  the  derivative  of 
another  function. 

The  derivative  of  a  function  may  be  defined  as  the  limit  of 
the  ratio  of  the  increment  of  the  function  to  the  increment  of 
the  independent  variable  when  the  increment  of  the  variable 
has  zero  for  its  limit. 

In  symbols  this  definition  may  be  formulated  as  follows. 
Let  the  function  be 

V  =  /(*)• 

Then  Lim 


if  it  has  a  limit,  is  the  derivative  of  the  function  /(x)  with 
respect  to  x.     The  steps  in  calculating  the  derivative  are: 

*-/(*). 

(1)  2/  +  Ai/=/(x  +  Ax). 

(2)  Subtracting,        A?/  =  f(x  +  Ax)  -/(x). 

/<>\   T^-  -j-      u     A     ty     f(x  +  Ax)  —f(x) 

(3)  Dividing  by  Ax,  ^|  =  -  ^        —• 

(4)  Taking  the  limit, 

&-  Urn         -  Eh.  *' 


The  symbols  -p,  Lim  rr*,  /'(x),  Dxy,  -r-  f(x)  are  to  be  used 

UX    Ax  —  »()iA.C  ('.(' 

synonymously  in  this  course  as  indicating  the  derivative  of  y 
with  respect  to  x,  where  y  =  /(x),  or  as  the  derivative  of  /(x) 
with  respect  to  x,  where  y  =  /(x). 

1.   Calculate    the    derivative   of  /(x)  =  x3  +  2  x  +  1    with 
respect  to  x.    Write 


196  LIMITS  AND  DERIVATIVES 

Giving  x,  y  increments,  Ax,  Ay  respectively 
y  +  Ay  =  (re  +  Ax)3  +  2(x  +  Ax)  +  1 

=  x3  +  3  x2Ax  +  3  x(Ax)2  +  (Ax)3  +  2  x  +  2  Ax  +  1. 

Subtracting,         Ay  =  3  x2Ax  +  3  xAx2  +  Ax3  +  2  Ax. 
Dividing  by  Ax,  ^  =  3  x2  +  3  xAx  +  (Ax)3  +  2. 

Taking  the  limit,       =  3  x2  +  2, 


since  all  other  terms  become  zero  when  Ax  —  »  0,  124. 

2.   Calcula 
to  x.    Write 


2    -£ 

2.  Calculate  the  derivative  of /(x)  =  re  —  ..    ,     ,  with  respect 

X     ^         M/ 


2x 
=  x  - 


1+x 

A         2  (x  +  Ax) 


A  2  Ax 

=  Ax  —  77—; 


Ax  (1  +  x  +  Ax)  (1  +  x) 


dx  (1  +  x)2 

3.  Calculate  the  derivative  with  respect  to  x  of 

y  =  3x3-5x2  +  2x-l. 

4.  Calculate  the  derivative  with  respect  to  x  of 

y  =  2  -  x  +  2  x2  -  3  x3. 

5.  Calculate  the  derivative  with  respect  to  x  of 


6.   Calculate  the  derivative  with  respect  to  x  of 

1 


DERIVATIVES  197 

7.   Calculate  the  derivative  with  respect  to  x  of 


8.  Calculate  the  derivative  with  respect  to  x  of 

y=(x-  3)4. 

Note.  —  Expand  first. 

9.  Calculate  the  derivative  with  respect  to  x  of 

y  =  x*  +  ;js  +  6- 

10.  Calculate  the  derivative  with  respect  to  x  of 

y  =  4  x2  —  z  —  :  -- 
1+a: 

128.  The  above  method  of  calculating  derivatives  is  general 
but  often  difficult  or  laborious.     To  facilitate  the  calculation  of 
derivatives,  several  special  rules  are  in  common  use.     These 
rules  enable  one  to  calculate  with  ease  the  derivatives  of  most 
ordinary  functions. 

129.  The  derivative  of  a  constant  is  zero.  —  This  follows 
from  the  fact  that  a  constant  can  have  but  one  value  and 
therefore  cannot  admit  an  increment.     That  is,  the  increment 
of  a  constant  is  zero  and  consequently  its  derivative  is  zero. 
This  may  be  symbolized  as 

(1)  |C  =  0, 

where  C  is  any  constant  and  x  any  variable. 

130.  The  derivative  of  x  with  respect  to  a;  is  1.     For  write 

x  =  x. 

Then  As  =  Az, 

A* 
As 

Arc 
and  Lim  -r—  =  1. 

(la) 


198  LIMITS  AND  DERIVATIVES 

131.  Derivative  of  ure,  where  u  is  a  function  of  x  and  n  a 
positive  integer.  —  Write 

y  =  un. 

y  +  Ay  =  (w  +  Aw) n  =  wn  +  nun~l  Aw  +  n  ^n  ~  ^  wn~2Aw2 * 

L? 

+  (terms  containing  Aw3  or  higher  powers), 

Ay  =  nw"-1  Aw  +  n(n~l\"-z  Aw2 
Lf 

+  (terms  containing  Aw3  or  higher  powers), 
Ay  ,  Aw  .  n(n  —  1)       „  A       Aw 

-T-*  =  mi71"1  -T \-      ^    tn '-  Wn~2  AW  •  -T— 

Ax  Az  |_2  Ax 

+  (terms  containing  Aw2  or  higher  powers). 

(2)  .-.      ife-n,^* 

*/.\  f/.\ 

since  all  other  terms  have  zero  for  a  limit  as  Az  — >  0,  124. 

This  equation  is  true  for  all  values  of  n,  fractional,  negative, 
irrational  or  imaginary.  If  desired  the  proofs  in  ^32,  133,  134, 
135  may  be  omitted. 

132.  Derivative  of  un  when  n  is  a  positive  fraction,  say 
n  =  p/q  where  p  and  q  are  positive  integers.     Write 

y  =  up/q. 

yq  =  up     (Raising  to  qih  power 


dy ln  up~ldu 

dx 


dx 

133.  Derivative  of  un  when  n  is  negative.  —  Write 

1 


y  +  Ay  = 


(w  +  Aw)m 
*[n««l«2»8»»»n,  read  factorial  n. 


Ay  = 


DERIVATIVES  199 

1  1 


(u  +  Aw)"1      um 

um  —  (um  +  mum~l  Aw  +  m    |  ~  -^wm~2Aw2  +  •  •  •  +  Aw 

(u-\-  Aw)mwm 

/         ,Aw      m(m  —  1)       9A      Aw  .  .    .        ,Aw 

Aj,  _  -I*""     AS  +  ~  J—  *      AuTx+-  '  '  +  AM      A^ 

Ax  w 


134.  Derivative  of  un  when  n  is  irrational.  —  Let  m  be  a 

variable  assuming  only  rational  values  as  it  approaches  the 
irrational  number  n  as  a  limit.  Write  n  =  m  -\-  e,  where 
e  —  >  0,  and  consider 

y  =  un  =  um+e  =  um  •  ue, 

where  e  is  independent  of  x.  Therefore,  since  we  —  >  1  indepen- 
dently of  x  we  have 

dy       d  /,.  \       d 

-T-  =  -r-[  lim  UmUe      =  -r-Un. 

ax      az\e__»o        /      dx 

(5)  /.      ^  =  nu—1, 

dx 

since  m  —  >  n  as  c  —  >  0. 

135.  Derivative  of  un  when  n  is  imaginary.  —  Let  n  =  mi, 
where  iz  =  —1.     Evidently  if  i  is  a  number  it  is  constant  and 
so  far  as  calculations  are  concerned  offers  no  new  principle. 
Hence  we  may  write  : 

y  =  umi. 

dy  mi    ,  du 

(6)  -r-  =  miu"11-1^— 
dx  dx 

Equations  2,  3,  4,  5,  6  show  that  the  equation  (2)  can  be  used 
for  all  values  of  n. 

136.  Derivative  of  the  product  of  two  or  more  functions.  — 
Write 

y  =  uv, 


200  LIMITS  AND  DERIVATIVES 

where  u  and  v  are  functions  of  x.    Then 
y  +  Ay  =  (u  +  Aw)  (v  +  At?) 


=  uv  +  wA0  +  »Aw  +  Aw  A0. 
Ay  =  wA0  +  0Aw  +  Aw  Az>. 
Ay  _      Ay        Aw       .      AM 
As  ~     Ax        AOJ  As 

,_,  dy          dt>  .      du 

(7)  -T-=u^r  +  v-r' 

dy          dx          dx 

Exercise.  —  Apply  formula  (7)  to  show  that  the  derivative  of 
a  constant  times  a  function  is  the  constant  times  the  derivative 
of  the  function.  That  is, 

(la)  S«*)-c£- 

137.  Derivative  of  the  quotient  of  one  function  divided  by 
another.  —  Write  ai 

y  =  u/v. 

Where  w  and  v  are  functions  of  x: 

Aw 


y  +  AT/  = 


v  +  Ac 

w  +  Aw      w 


Aw         A0 

A  ^1  --  W-r— 

Ay  _     Ax        Ax 

Ax        (v  +  Av)  0 

du          dv 

,        v-j  --  WT- 

dy          <iag          dx 


138.  Derivative   obtained   from   an   implicit   function. 

This  method  will  be  illustrated  by  examples.    Consider 


DERIVATIVE  OBTAINED  FROM  AN  IMPLICIT  FUNCTION    201 
Apply  the  formulas  of  129,  130,  131  to  each  term.    The  result 


Solving  for  -g, 

(a)  ^  -  — 

dx  ~    a?y  ' 

since  -r-  =  1.    By  use  of  the  original'  equation  y  may  be  elimi- 
dx 

nated  from  this  derivative,  giving  the  result  in  terms  of  x. 
Again  consider 

axy  +  6  =  0. 

Applying  the  above  formulas  to  each  term, 


Hence 

<w  dJ.=  -y. 

dx          x 
Find  the  derivatives  of  the  following: 

1.  y  =  3z4-6z+l  (formulas  (1)  and  (2)). 

2.  y  =  2x(x2  +  1)  (as  a  product,  u  =  2  x,  v  =  (x2  -f-  1)  or 
write  2  y?  +  2  x. 

3.  y  =  (x  —  2)  (x2  +  4z)  (as   a   product   w  =  (x  —  2),   »  = 
(x2  +  4  x)  or  multiply  and  use  (1)  and  (2). 

4.  y  =  -^-  (formula  (8)). 

x  -\-  1 

5.  i/2  +  2  z?/  -  x2  =  10  (implicit  function  138). 
6-  V  =  2  (f°nnulas  (2)  and  (8)). 


8.  ?/  =  (1  —  z3)^  w^  where  w  =  1  —  x3. 

9.  xY  +  z2  +  2/2  +  1  =  0. 


202  LIMITS  AND  DERIVATIVES 

A  derivative  has,  in  general,  a  definite  value  for  an  assigned 
value  of  the  variable.  This  value  is  found  by  substituting  the 
value  of  the  variable  in  the  derivative. 

10.  y*  =  6  x3  +  4,  find  ^  for  x  =  6,  and  x  =  0. 

ax 

11.  x2  +  y*  =  25,  find  j-  for  x  at  the  point  (3,  4). 

dii 

12.  xy  =  12,  find  ~  for  x  =  12.     Find  y  from  the  equation. 

13.  x/y  +  1/x2  =  y/x  +  y,  find  dy/dx  for  a;  =  1. 

14.  y  =  (a2  -  x2)*,  find  ^  for  3  =  a. 

15.  £/  =  -        —  j-  find  ~  f  or  x  =  2,  and  x  =  0. 

(1  —  a?)*          «£ 

16.  For  what  values  of  x  is  the  derivative  of  x2  —  x  -f-  5 
equal  to  0?     Equal  to  10?     (Equate  the  derivative  to  0  and 
solve  for  x.     Similarly,  for  all  values.) 

17.  For  what  values  of  x  is  the  derivative  y  =  x3  —  9  x  equal 
to  0?    Equal  to  4? 

dA 

18.  The  area  of  a  circle  is  A  =  irr2.     Find  -r-  when  r  =  12. 

dr 

19.  From  xy  =  12,  find  dy/dx  when  x  =  1,  x  =  12. 

20.  From  y"-  =  4  px,  find  dy/dx  when  x  =  p. 


139.  Use  of  the  derivative  to  find  the  slope  of  a  curve.  - 

The  slope  of  a  straight  line  was  defined  in  28.     The  slope  of  a 
curve  at  a  point  is  the  slope  of  the  tangent  line  at  that  point. 
Let 

(1)  */=/(*) 

be  the  equation  of  any  curve  and  let  PI  (xi,  3/1)  be  any  given 
point  on  the  curve  and  P  (x,  y)  a  variable  point  in  the  neighbor- 
hood of  Pi  (xi,  yi).  The  slope  of  the  chord  PiP  is 

Ayi  _  y-yi  _ 

i  —  =  -  =  tan  p. 

Axi      x  —  Xi 


EQUATION  OF  TANGENT 


203 


As  P  moves  toward  Pi  as  a  limiting  position,  the  chord  PiP 
approaches  the  position  of  TPi,  the  tangent  at  PI.     Then  also 


=  -tan*. 


FIG.  99. 

It  follows  that  the  slope  of  a  curve  at  a  given  point  and  the 
slope  of  the  tangent  to  the  curve  at  that  point  are  each  equal 

di/ 
to  the  derivative  -j-  at  that  point. 

CM/ 

It  is  now  easy  to  obtain  the  equation  of  the  tangent  line  to  a 
curve  at  a  given  point,  if  the  equation  of  the  curve  and  the 
coordinates  of  the  point  are  known.  By  99,  Eq.  3,  the  equation 
of  the  line  having  a  given  slope  and  passing  through  a  given 
point  is 

y  —  2/1  =  m  (x  —  Zi). 


The  notation 


means  that  x\  is  substituted  for  x  in  the  deriva- 


tive, and  t/i  for  y  also  if  y  occurs  in  the  derivative. 


204  LIMITS  AND  DERIVATIVES 

For  a  line  tangent  to  y  =  }(x)  at  the  point  (xi,  yi)  we  must, 
therefore,  have 

(9)  y  -  yi  =  \ 

This  is  the  equation  of  the  tangent  line  at  (xi,  y\). 

1.  Find  the  equation  of  the  tangent  to  yz  =  8  x  at  the  point 
where  x  =  8. 

dy      4 

Here  -/-  =  — 

ax     y 

By  use  of  the  equation  of  the  curve,  y  =  8  when  x  =  8.     Hence 

dy~]       _  4~|       _  1^ 
dx]x=s     y]y  =  s      2 

Then  y  -  8  =  %  (x  -  8) 

or  2y  —  x  =  8 

is  the  desired  equation  of  the  tangent  line. 

x2 

2.  Find  the  equation  of  the  tangent  to  y  =  -^  at  the  point 

where  x  =  2. 

3.  Find  the  equation  of  the  tangent  to  2  ?/2  —  xz  =  4  at  the 
point  where  x  =  4. 

4.  Find  the  equation  of  the  tangent  to  x2  +  y2  =  25  at  the 
point  where  x  =  3. 

5.  Find  the  equation  of  the  tangent  to  y  =  x3  at  the  point 
where  x  =  —  1. 

6.  At  what  angle  do  x2  +  yz  =  25  and  xy  =  12  intersect  ? 
./Vote.  —  The  angle  between  two  curves  at  their  point  of 

intersection  is  the  angle  between  their  tangents  at  the  point  of 
intersection.     See  102. 

7.  At  what  angle  do  x*  —  y2  =  36  and  2  x  —  3  y  =  1  inter- 
sect? 

8.  Show  that  y2  =  4  px  has  two  tangents  for  x  =  p  and  that 
these  tangents  meet  each  other  at  right  angles  on  the  z-axis. 

9.  Show  that  the  tangent  line  at  any  point  of  the  parabola 
y*  =  4  px  makes  equal  angles  with  the  line  from  the  focus  to 


MAXIMUM  AND  MINIMUM  VALUES  205 

the  point  of  contact  and  a  line  through  that  point  parallel  to 
the  x-axis.  What  practical  use  is  made  of  this  property  ? 

10.  Show  that  the  tangent  at  any  point  of  the  ellipse  makes 
equal  angles  with  the  lines  from  the  foci  to  the  point  of  contact. 

140.  Maximum  and  minimum  values  of  functions  are  of 
frequent  occurrence  and  of  much  importance.  As  a  simple 
example,  consider  the  case  of  a  body  thrown  vertically  upward, 
to  find  the  greatest  height  it  will  ascend  when  its  initial  velocity 
is  given.  The  law  of  the  falling  body  must  be  combined  with 
the  law  of  uniform  motion.  The  formula  is 

s  =  v0t  -  \  gt2, 

where  s  is  the  height,  VQ  the  initial  velocity,  g  the  acceleration 
of  gravity  and  t  the  time  in  seconds  from  starting. 
Solving  the  equation  for  t  gives 


_  VQ  ±      v0z  —  2gs 
t  —  -  • 

g 

Since  v0,  g,  s  are  essentially  positive  and  t  must  be  real  the 
expression  v<?  —  2  gs  must  not  be  negative.  Hence  s  may 
become  so  large  as  to  make  v02  —  2  gs  =  0  but  cannot  increase 
further  without  making  t  imaginary.  Hence  solving  v02  —  2  gs 
=  0  for  s  gives  the  maximum  value  of  s  to  be  s  =  v02/2  g. 
This  method  is  not  easy  to  carry  out  in  most  cases  that  arise  in 
scientific  investigations.  A  much  more  powerful  and  general 
method  is  given  below. 

141.  Use  of  the  derivative  to  determine  maximum  and 
minimum  values  of  functions.  —  All  ordinary  functions  cf 
one  variable  admit  of  graphic  representation.  For  this  reason 
geometrical  problems  are  convenient  and  sufficient  to  illustrate 
the  use  of  the  methods.  These  can  be  immediately  transferred 
to  any  field  of  science  by  the  medium  of  graphic  representation 
which  is  of  almost  universal  application. 

Definitions.  —  A  maximum  ordinate  of  a  curve  is  one  which 
is  algebraically  greater  than  ordinates  on  both  sides  of  itself 
however  near  these  ordinates  be  chosen. 


206 


LIMITS  AND  DERIVATIVES 


A  minimum  ordinate  of  a  curve  is  one  that  is  algebraically 
less  than  ordinates  on  both  sides  of  itself,  however  near  they 
are  chosen. 

In  Fig.  100,  y  and  y"  are 
maximum  ordinates  and  P 
and  P"  are  called  maximum 
x  points  of  the  curve.  Simi- 
larly, y'  and  y'"  are  minimum 
ordinates. 

Let  y  =  f(x)  be  any  con- 
tinuous  function   of   x  and 
FIG.  100.  suppose   that  its  derivative 

is  continuous  and  one-valued  in  the  region  considered.    Then 
if  2/1  =  f(xi)  is  &  maximum,  we  shall  have 
(1)  /(*i  ~  h)  </(#i)  > /(#i  +  h), 

however  small  h  is  chosen.     Similarly,  for  a  minimum, 

It  follows  from  the  continuity  of  the  derivative  and  the 
definition  of  maximum  and  minimum,  that  if  x\  corresponds  to 

dii 
a  maximum  of  the  function,  the  derivative  ~r  =  f'(x)  will  be 

ft  T 

positive  for  values  of  x  in 
the  interval  x\  —  h  to  Xi, 
negative  in  the  interval  x\ 
to  x\  +  h  and  zero  for  x  = 
Xi]  this  is  easily  seen  from 
Fig.  101.  For  evidently 
the  slope  of  the  tangent  is 
positive  along  the  arc  APi, 
negative  along  the  arc  P\B 
and  zero  at  PI.  Similarly  if 
x\  corresponds  to  a  mini-  FIG.  101. 

di/ 
mum  the  derivative  •?•  =  f'(x)  will  be  negative  for  values  of  x 

in  the  interval  Xi  —  h  to  x\.  positive  in  the  interval  Xi  to  Xi  +  h 


INCREASING  AND  DECREASING  FUNCTIONS         207 

and  zero  for  x  =  x\.  In  the  figure  it  is  seen  that  the  slope  of 
the  tangent  is  negative  along  the  arc  A'P2,  positive  along  the 
arc  P2B'  and  zero  at  P*. 

A  necessary  condition  that  y  =  f  (x)  shall  be  a  maximum  or 
a  minimum  value  at  x\  is  that 


It  must  be  noted  that  this  condition  is  not  sufficient  for  it  may 
hold  at  points  where  the  function  is  neither  a  maximum  nor  a 
minimum  as  can  be  seen  in  the  annexed  figure. 

It  is  now  necessary  to  determine 
when  the  condition  above  gives  a  max- 
imum, a  minimum  or  neither.  This 
can  be  done  by  the  use  of  the  inequal- 
ities  (1)  and  (2).  The  steps  for  deter-  ' 
mining  maximum  and  minimum  values 
of  a  function  are: 

1.  Calculate  the  derivative  with  re-  FlG-  102- 

spect  to  the  variable,  and  eliminate  the  dependent  variable  if 
it  occurs. 

2.  Equate  the  derivative  to  zero,  giving  f'(x)  =  0,  if  x  is  the 
variable. 

3.  Solve  the  equation  in  (2).    Call  its  roots  critical  values 
of  the  variable.     Denote  them  by  Xi,  x%,  .  .  .  ,  xn. 

4.  To  determine  whether  any  one  of  these,  say  xi}  gives  a 
maximum  or  a  minimum,  choose  a  convenient  value  of  h  and 
apply  the  inequalities  (1),  (2). 

5.  Having  determined  which  critical  values  give  maximum 
and  which  minimum  values,  substitute  the  critical  values  in 
the  function  and  determine  the  actual  maximum  and  minimum 
values  of  the  function. 

142.  Consider  y  =  f(x),  a  function  of  x.  If  f(x)  increases 
in  value  when  x  increases  and  decreases  when  x  decreases.  f(x) 
is  called  an  increasing  function  of  x.  If  f(x)  decreases  when  x 
increases  and  increases  when  x  decreases,  f(x)  is  called  a  de- 
creasing function  of  x. 


208  LIMITS  AND  DERIVATIVES 

Since,  if  y  =  /(x)  is  a  decreasing  function  of  x,  the  increments 

di] 
A  i/  and  Ax  are  of  opposite  signs,  -p  =  /'(x)  is  negative.    A 

given  function  may  be  an  increasing  function  in  one  interval 
and  a  decreasing  function  in  another  interval. 
Thus  for  a  maximum : 

(1)  /'(*!   -  fc)   >  /'(*0    =   0  >  /' (*1  +  ti). 

For  a  minimum: 

(2)  /'(*!  -  Ji)  </' (*0  =  0  </'(*!  +  fc). 

This  method  is  quite  similar  in  manipulation  to  the  method  ex- 
plained in  141.  The  student  will  find  this  method  applicable 
to  many  problems  when  the  second  derivative  is  not  easily 
obtained.  (See  143  for  second  derivative.) 

Consider  y2  =  25  x2  —  x*,  and  find  the  maximum  value  of 
this  function.     Solving  for  y, 

f(x)  =  y  =  ±x  V25  -  x2, 

f'(x)  —  -T-  =    /^          —  0  (use  positive  radical), 


/'(*)] 
Ji 

H. 


V2 
25-2 


v  25  -  32 
25  -  2  -  42 


=  ±  3.54, 

*2  =  +  (/i  =  .54) 

=  -.  (ft  =  .46) 


V25  -  42 

Hence/' (x)  changes  sign  from  +  to  —  at  the  point  x  =  3.54  + 
and  /(x)  is  a  maximum  for  this  value  of  x.  Substitute  x  = 
3.54  in/(x)  and  determine  the  maximum  value. 

Let  the  student  test  the  negative  value  of  x  =  —3.54.  Also 
solve  the  problem  using  the  negative  radical.  Draw  a  graph 
of  the  function  and  discuss  this  curve. 

143.  In  general  the  derivative  /'(x)  of  /(x)  is  itself  a  func- 
tion of  x  and  will  have  a  derivative  with  respect  to  x.  The 
derivative  of /'(x)  is  the  second  derivative  of /(x)  and  is  denoted 

by  the  symbols  -|,  -7-5,  /(x)  or/"(x). 


MAXIMUM  AND  MINIMUM  VALUES  209 

In  the  neighborhood  of  a  maximum  point  the  function  f(x') 
of  y  =  f{x)  was  shown  to  be  first  positive,  then  zero  and  nega- 
tive for  increasing  values  of  x.  It  is,  therefore,  a  decreasing 
function  of  x.  Therefore  if  Xi  corresponds  to  a  maximum  value 
°f  2/  —  /(x)  we  must  have 

/'(%)  =  0 
and 

f'M  <  0. 
At  a  minimum  point  we  must  have  in  a  similar  way 

/'(*,)  =  0. 
/"(*i)  >  0. 

These  inequalities  and  equations  can  be  used  for  determining 
maximum  and  minimum  values  of  functions  instead  of  the 
methods  of  141,  142. 

Some  illustrative  examples  will  now  be  given. 
1.   Solve  the  problem  at  the  beginning  of  140  by  use  of  the 
derivative.     Write 

s  =  VQ[,  —  |  gP. 

Calculate  the  derivative  of  s  with  respect  to  t, 

ds 

#  =  *  -  9t- 

For  a  maximum  this  derivative  must  be  zero.    Hence 

v0  -  gt  =  0, 
t  =  v0/g. 

This  is  the  critical  value  of  t.    Calculate  the  second  derivative 


This  is  negative  for  all  values  of  t,  since  g  is  independent  of  t. 
The  function,  therefore,  has  a  maximum  for  t  =  v0/g.  Sub- 
stituting this  value  of  t  in  the  function  gives 


which  is  the  same  result  as  obtained  before. 


210  LIMITS  AND  DERIVATIVES 

Instead  of  using  the  second  derivative  we  may  choose  a  value 
of  h  and  try  to  satisfy  inequality  (1)  141.     Thus,  let  h  =  e,  a 

small  number,  and  substitute  t  =  —  —  e  and  t  =  —  +  e  in 

9  Q 

the  original  function  and  compare  the  result  with  that  for 

t  =  v0/g. 


0  20 

The  last  value  is  greater  than  the  first  two  for  all  values  of  c, 
however  small.  Then  the  last  value  of  s  is  a  maximum.  This 
solution  illustrates  both  methods  of  procedure. 

2.  Consider  the  problem:  To  prove  that  of  all  rectangles 
having  a  given  area  the  square  has  a  minimum  perimeter. 
Let  A  =  xy,  where  x,  y  are  the  length  and  breadth  of  the 
rectangle  respectively.  The  perimeter  is 

P  =  2x  +  2y, 

dp  -9  +  9dy 

dx~  =          2dx' 

2  +  2^  =  0. 
dx 

dij 

To  eliminate  -?•  take  the  derivative  of  A  =  xyt 
dx 

dA  dy 

-=y  +  x-=0, 

since  A  is  constant. 

dy  _  _  y. 

dx          x 
Substituting  in  -T-  =  0  gives 

2  -  2  y/x  =  0 
or  x  =  y. 


MAXIMUM  AND  MINIMUM  VALUES  211 

That  is  the  length  is  equal  to  the  breadth  and  the  rectangle  is 
a  square.     From  A  =  xy  and  x  =  y 

x*  =  y*  =  A 

and  x  =  y  =  VA, 

critical  value  of  x.    Choose  x  =  VA  —  h,  whence  by  A  =  xy, 

A 

y  =  —p=  -- 
VA  -  h 

Hence      P  =  2x  +  2y  =  2  \VI  -h  +  -^  -  1  >  4  VZ. 

L  v  A  —  hJ 

Choose  x  =  VA  +  h,  whence  by  A  =  xy, 

A 

~  Vl+fc* 

Hence  P  =  2x  +  2y  =  2  |~VZ  +  h  +  -7=^  —  1  >  4  VI. 

L  V  A  +  hj 

These  results  show  that  if  x  9^  y  the  perimeter  is  greater  than 
when  x  =  y  and  the  minimum  value  of  P  is  4  VZ. 
To  apply  the  method  of  143  we  have 


dxz         dx2 

A  d?y      2A 

==x'  W  =  ^' 

Substituting  the  critical  value,  x  =  +  \/A,  gives 

d2P      4A1  4A 

-5-5-  =  —  T-  =  —7=  >  0,  since  A  >  0. 

dx2       x3  ]Z=VA      VA3 

This  test  shows  that  the  value  x  =  VA  makes  P  a  minimum 
and  consequently  from  the  condition  above  x  =  y,  the  rectangle 
is  a  square.  Both  methods  lead  to  the  same  conclusion. 

3.  What  are  the  dimensions  of  a  tomato  can  of  capacity 
63  cu.  in.  of  such  form  as  to  require  a  minimum  of  material, 
no  allowance  being  made  for  seams? 

Write 
(1)  V  =  itfh  =  63, 


212  LIMITS  AND  DERIVATIVES 

where  r  is  the  radius  and  h  the  height  of  the  can.    Also, 

(2)  A=2Trr2  +  2irrh, 

where  A  is  the  area  of  the  total  surface  of  the  can.    Then 

dA  ,       0      dh      _ 

-j-  =  4  TIT  +  2irh  +  2irr-r=  0, 
dr  dr 

for  a  minimum.    This  becomes 

(3)  2r  +  h  +  r^  =  Q. 

From  (2), 

dfc          2  A 

(3a)  -3-  =  --- 

dr  r 

Substituting  in  (3)     2r  +  /i-2fc  =  0 
or  2  r  =  h. 

From  (1)  by  substitution  2  irr3  =  63. 
Whence  r  =  2.15  (critical  value  of  r) 

and  ft  =  4.30. 

From  (3), 

...  d2A  dh.-     d% 

(4)  -j-r  =  4ir  +  47r-7-  +  2irr-7v 
dr2  dr  dr2 

,,  N  d2^      6/1 

From  (3a),  ^  =  -r 

Substituting  values  of  -T-,  -ri,  r  and  /i  in  (4), 


...  dM. 

(5)  ^  =  47r-  r 

and  a  minimum  of  A  is  indicated. 
Substituting  values  of  r  and  h  in  (2)  gives 
A  =87.1  sq.  in. 

as  the  minimum  value  of  A. 

By  substituting  r  =  2,  r  =  2.3  for  r  in  the  expression  for  A, 
remembering  h  =  2  r,  gives  for  A,  respectively,  the  values 

A  =88 
and  A  =  88  —  , 


MAXIMUM  AND  MINIMUM  VALUES  213 

both  of  which  are  greater  than  the  value  obtained  above. 
Thus  again  the  two  methods  agree  in  their  results. 

Remark.  —  Hereafter  the  student  may  solve  a  problem  by  one 
method.  He  may  choose  which  method  to  use.  In  general  if  the 
second  derivative  is  easily  obtained  that  method  will  be  best. 

1.  Show  that  of  all  rectangles  of  given  perimeter  the  square 
has  a  maximum  area. 

2.  Show  that  of  all  rectangles  inscribed  in  a  circle,  the  square 
has  a  maximum  area. 

Note.  —  Assume  the  radius  of  the  circle  equal  to  r.  Consider 
the  geometric  properties  of  the  figure  and  form  an  expression 
for  the  area  of  the  rectangle  using  a  variable  dimension  and 
solve  the  problem. 

3.  Show  that  of  all  triangles  of  a  given  base  and  perimeter 
the  isoceles  triangle  has  the  maximum  area. 

4.  Find  the  dimensions  of  a  cylinder  of  maximum  volume 
that  can  be  inscribed  in  a  cone  of  radius  10  and  altitude  20. 

5.  Find  the  dimensions  of  a  cone  of  volume  3000  cu.  ft.  that 
shall  have  a  minimum  curved  surface. 

6.  A  fireplace  is  2'  deep  and  4'  high.     Find  the  length  of  the 
longest  straight  pole  that  can  be  pushed  up  the  1'  chimney. 

7.  Find  the  lowest  point  (minimum  ordinate)  of  the  curve 
y  =  x2  +  x  -  6. 

x  -\-  6 

8.  Does  the  curve  y  =  have  maximum  or  minimum 

ordinate? 

9.  A  carpenter  has  108  sq.  ft.  of  lumber.     Find  the  dimen- 
sions of  the  box  of  maximum  capacity  (lid  included)  that  he 
can  make,  making  no  allowance  for  joining. 

10.  A  rectangular  sheet   of  iron   is   12"  x  18".     Find  the 
dimensions  of  the  rectangular  pan  of  maximum  capacity  that 
can  be  made  by  folding  up  the  edges. 

11.  Find  the  legs  of  the  right  triangle  of  maximum  area  that 
can  be  constructed  on  a  hypotenuse  of  24". 

12.  Find  the  dimensions  of  a  right  circular  cylinder  of  maxi- 
mum volume  that  can  be  inscribed  in  a  sphere  of  radius  12. 


214  LIMITS  AND  DERIVATIVES 

13.  In  problem  5,  28,  use  the  three  points  corresponding  to 
the  prices  $12,  $18,  $24,  of  the  profit  curve  and  determine  an 
equation  of  the  form  y  =  axz  +  bx  +  c.  From  this  equation 
determine  the  maximum  profit  and  the  price  corresponding. 

144.  Use  of  the  derivative  to  define  motion.  —  We  are 
familiar  with  "speed  of  a  train"  or  "rate  of  walking"  of  a  per- 
son. It  is  customary  to  say  the  speed  of  a  moving  body  is  the 
distance  traveled  divided  by  the  time  required  to  travel  the 
distance.  This  is  only  an  approximate  expression  of  the  idea. 
What  is  meant  is  —  the  number  of  units  distance  covered, 
divided  by  the  number  of  units  of  time  required,  is  the  numerical 
measure  of  the  average  speed  over  the  distance  during  the 
interval  of  time. 

If  speed  is  not  uniform  during  the  interval  it  will  be  more 
than  the  average  speed  during  part  of  the  interval  and  less  than 
the  average  during  a  part  of  the  interval.  If,  at  a  given  instant 
of  time,  the  speed  of  a  body  becomes  uniform,  then  the  distance 
As,  passed  over  in  the  time  A£,  is  the  instantaneous  speed  at 
the  given  instant.  We  shall  write,  therefore, 

Instantaneous  speed  =  -r-» 

under  the  above  assumptions. 

Consider  the  example  of  a  train  moving  as  follows: 
The  train  moves  40  mi.  during  1  hr. 

"    .  "        "      25    "        "      the  first    |  hr. 

It  ((  II  10         <(  "  <(  i         ft 

"      "        "        5    "       "  "          5min. 

it        tt  tt        11     ..  u  ((  i       « 

The  average  speed  for  each  interval  is  as  follows: 
For     1  hr.  40  mi.  per  hr. 

"       first     Jhr.  50    " 

"         "       \  "  52    "        " 

"       5min.  60    " 

ti         tt       i     n  gg    it        tt 

Which  of  the  above  values  of  the  speed  is  probably  nearest  the 
actual  speed  at  the  beginning  of  the  hour? 


USE  OF  THE  DERIVATIVE  TO  DEFINE  MOTION      215 

Suppose  the  interval  of  time  be  decreased  toward  zero  as  a 
limit.     The  speed  will  also  approach  some  value  as  a  limit. 

As 

The  limiting  value  of  the  ratio  -rr  ,  as  Ai  —  »  0,  is  the  instan- 

taneous speed  at  the  beginning  of  the  interval,  A£.  But  we 
know  that  if  s  is  some  function  of  t,  the  limit 

,.      As      ds 

htn  -TT  =  37 

At—^o  Ac      at 

is  the  derivative  of  s  with  respect  to  t.  That  is,  the  instantane- 
ous speed  is  the  derivative  of  s  with  respect  to  t,  when  s  is 
regarded  as  a  function  of  t.  It  is  noted  that  so  far  as  the 
mathematician  is  concerned  s  and  t  are  merely  two  variables 
related  in  some  way  and  that  he  may  with  equal  propriety 
regard  dy/dx  as  the  instantaneous  speed  of  y  with  regard  to  x, 
or  re-rate  of  y. 

Linear  acceleration  is  defined  as  the  rate  (speed)  of  change  of 
speed.     Therefore  linear  acceleration  may  be  written 

do    .  d  fds\      cPs 
acceleration  =      =  =      , 


where  s  is  regarded  as  a  function  of  tune,  I.    . 

1.  A  circular  plate  6"  in  diameter  is  expanding  by  heat  so 
that  its  radius  is  increasing  at  the  rate  of  1"  per  sec.  At  what 
rate  is  the  area  of  the  plate  increasing? 

Write 
(1)  A=irr>. 


rru 

Then  -3r  J 

at  at 

dr 
Substituting  r  =  3  and  37  =  1  from  the  problem, 

dA 

-57-  =  6  T  sq.  in.  per  sec. 

2.  The  edge  of  a  cube  is  12"  and  is  increasing  at  the  rate  of 
2"  per  sec.  At  what  rate  is  the  volume  increasing?  The 
surface? 


216  LIMITS  AND  DERIVATIVES 

3.  At  what  rate  is  the  area  of  a  rectangle  increasing  if  its 
sides  a,  b  are  each  increasing  at  the  rate  of  c"  per  sec.  ? 

4.  Water  runs  into  a  conical  vessel  at  the  rate  of  3  cu.  in.  per 
sec.    The  diameter  of  the  top  is  12",  the  altitude  is  18",  vertex 
downward.     At  what  rate  is  the  depth  of  the  water  increasing 
when  it  is  8"  deep  in  the  vessel? 

Note.  —  Consider  a  cylinder  of  height  unknown  and  radius 
equal  to  the  radius  of  the  vessel  at  8"  from  the  vertex  and  of 
volume  3  cu.  in. 

5.  A  man  6'  tall  walks  along  a  path  which  passes  under  a 
light  12'  above  the  path.     He  walks  at  rate  of  3  mi.  per  hr. 
from  the  light.     At  what  rate  is  his  shadow  changing  length 
when  he  is  25'  from  a  point  under  the  light? 

6.  A  body  moves  so  that  its  distance  from  a  fixed  point  is 
given  by  s  =  ts  +  6  t2  -  10  x  +  18. 

(a)  What  is  the  speed  when  t  =  4?    When  t  =  0? 
(6)  What  is  the  distance  when  t  =  4?    When  t  =  5? 

(c)  What  is  the  acceleration  when  t  =  10  ?    When  t  —  1  ? 

(d)  Discuss  the  graph  of  each  of  the  functions  a,  b,  c,  using 
t  as  abscissa. 

7.  Construct  on  the  same  axes  the  graphs  of  y  =  t3  and  of 
the  speed  and  the  acceleration  of  this  function. 

8.  Construct  the  graphs  of  s  =  16  •  t2  and  of  its  speed  and 
acceleration  functions. 

9.  A  particle  moves  in  a  path  so  that  the  coordinates  of  its 
position  are  x  =  1  —  t  +  t2,  y  =  1+  t  +  t2.     Show  that  the 
path  is  a  parabola  when  referred  to  x,  y  coordinates  and  that 
the  speed  of  the  particle  is  an  increasing  function  of  t. 

10.  The  electrical  resistance  of  platinum  wire  varies  as  its 
temperature  6°  C.,  according  to  the  law  R  =  R0  (1  +  aB  +  &02)"1. 

jr> 

Calculate  -^  and  interpret  the  meaning  of  this  derivative. 

uu 

11.  A  body  moves  on  y2  =  12  x.     At  what  rate  is  it  moving 
parallel  to  the  z-axis  if  it  moves  50'  per  sec.  in  the  path  when 
x=  10. 

Note. —  Calculate  the  component  of  speed  parallel  to  the  z-axis. 


EQUAL  ROOTS  OF  EQUATIONS         217 

145.  Equal  roots  of  equations.  —  From  85  it  is  evident 
any  integral  function  in  one  variable  (or  unknown)  can  be 
written  in  the  form 

f(x)  =  k(x  -  n)  (x  -  r2)  (x  -  r3)  .  .  .   (x  -  rn), 

where  A;  is  a  constant  and  n,  rz,  .  .  .  rn  are  the  roots  of  the 
equation, 

(1)  /(*)  =  0. 

If  m  of  these  roots  are  equal,  equation  (1)  may  be  put  in  the 
form, 

(2)  f(x)  =  k(x  -  n)m  (x  —  r2)  (x  -  r8)  .  .  .  (x  -  rre_m)  =  0. 
Call          k(x  -  rs)  (x  -  r3)  .  .  .  (x  —  rn_ro)  =  <f>  (x). 

Then  (2)  may  be  written 

(3)  /(*)  =  (*-r1)»0(x)=0. 
Now 

(4)  f'(x)  =  m(x  -  rO"1-1  <j>(x)  +  (x  -  ri)m4>'(x}  =  0. 

The  highest  common  factor  of  /(x)  and  /'(#)  contains  the 
factor  (x  —  ri)™"1.  The  equation, 

(5)  (x  -  n)"1'1  =  0, 

contains  r\  as  a  root  one  less  time  than  does  f(x)  =  0.     It  is 
easy  to  see  that  if  there  were  no  multiple  *  roots  in  f(x~)  =  0, 
there  could  be  no  highest  common  factor  containing  x. 
If  (3)  should  be  of  the  form, 

(6)  f(x)  =  k(x-  n)"  (x  -  r2) 1 0(a;)  =  0, 
then  corresponding  to  (5)  is 

(x  -n}m~l(x  -r2)z~1  =  0. 

The  reasoning  applies  for  any  number  of  multiple  roots.  Hence 
to  determine  whether  an  equation  has  multiple  roots  and  to 
determine  these  roots  when  they  exist,  we  may  proceed  as 
follows: 

(a)  Calculate  the  derivative  of  f(x). 

*  When  the  same  root  occurs  two  or  more  times  in  an  equation  it  is 
called  a  multiple  root. 


218  LIMITS  AND  DERIVATIVES 

(6)  Calculate  the  highest  common  factor  of  /  (x)  and  /'  (x) 
and  call  it  F(x). 

(c)  Solve  the  equation  F  (x)  =  0. 

(d)  Each  root  of  F  (x)  =  0  will  occur  in  f(x)  =  0  one  more 
time  than  in  F  (x)  =  0. 

1.  Determine  the  multiple  roots  of  x6  —  2  x2  +  2  x  —  2  =  0. 

2.  Determine  the  multiple  roots  of  4  x3  —  16  x2  +  52  x  —  3 
=  0. 

3.  Determine  the  multiple  roots  of  x4  —  16  =  0. 

4.  Determine  k  so  that  the  roots  of  x'*  —  6  x  +  A;  =  0  shall 
have  its  roots  equal. 

146.  Momentum  is  defined  as  the  product  of  mass  and 
velocity.  Velocity  being  a  vector,  momentum  is  a  vector. 
If  only  the  numerical  value  of  velocity  is  considered,  the  product 
of  speed  and  mass  is  the  numerical  value  of  momentum.  In 
this  sense  consider 
(1)  M  =  mo, 

where  m  is  mass  and  v  is  speed.    Taking  the  derivative 

dv  , 


since  mass  multiplied  by  the  magnitude  of  acceleration  is  the 
magnitude  of  force.  Equation  (2)  shows  that  force  is  the  time 
rate  of  change  of  momentum. 


CHAPTER  XVI 
SERIES;  TRANSCENDENTAL  FUNCTIONS 

147.  A  sequence  *  is  called  arithmetic  if  the  differences  be- 
tween successive  numbers  of  the  sequence  are  equal  through- 
out the  sequence.     In  other  words  when  every  number  of  the 
sequence  after  the  first  may  be  found  by  adding  the  same 
number  to  the  preceding  number  the  sequence  is  called  arith- 
metic.   The  number  added  to  a  number  of  an  arithmetic  se- 
quence to  obtain  the  succeeding  number  is  called  the  common 
difference.     Thus,  the  numbers 

2,     5,     8,     11,     14,     17 

form  an  arithmetic  sequence  whose  common  difference  is  3. 

A  sequence  is  called  geometric  if  each  number  after  the  first 
in  the  sequence  is  obtained  from  the  preceding  number  by 
multiplying  by  the  same  number.     The  multiplier  is  called  the 
common  ratio  of  the  sequence.     Thus,  the  numbers 
2,    4,     8,     16,     32,     64 

form  a  geometric  sequence  whose  common  ratio  is  2. 

The  law  connecting  successive  numbers  of  a  sequence  may 
be  simple  or  complicated.  When  this  law  can  be  expressed  in 
the  form  of  an  equation  it  is  called  a  recursion  formula. 

148.  If  ai,  02,  a3,  .  .  .  ,  an  is  a  sequence  of  numbers  of  any 
kind,  then 

(1)  ai  +  02  +  a3  +  •  •  •  +  an 

is  called  a  series.  If  n  is  infinite  the  series  is  called  an  infinite 
series.  The  numbers  a\,  0%,  a3  .  .  .  are  called  the  terms  of  the 
series. 

*  For  definition  of  sequence  see  37c. 
219 


220  SERIES;  TRANSCENDENTAL  FUNCTIONS 

The  reason  for  studying  series  lies  in  the  fact  that  a  number 
of  the  functions  with  which  we  have  to  do  in  mathematics  and 
its  applications  are  most  naturally  studied  by  means  of  the 
series  which  represent  them.  Many  problems  are  of  such  a 
nature  as  to  lead  quite  naturally  to  series  in  their  solution. 
For  purposes  of  this  course  we  shall  consider  only  one  class  of 
series,  viz.,  convergent  series. 

A  convergent  series  may  be  defined  as  follows  :  If 

(2)  Sn  =  ai  +  (h  +  a3  +  •  •  •  +  an 
is  the  sum  of  the  first  n  terms  of  the  series 

(3)  S  =  01  +  02  +  a3  +  •  •  •  +  a»  +  •  •  • 

and  if  Sn  approaches  a  definite  limit  as  n  increases  indefinitely 
the  series  S  is  said  to  be  convergent.  This  means  that  by 
taking  the  sum  of  a  sufficient  number  of  terms  of  a  convergent 
series  the  difference  between  this  sum  and  the  limit  of  that 
sum  may  be  made  as  small  as  one  pleases.  A  series  with  a 
finite  number  of  terms  is  always  convergent  if  the  terms  are 
finite.  In  (3)  S  =  lim  Sn  is  sometimes  called  the  sum  of  the 

n  —  *°o 

series.     S  will  also  be  used  to  denote  the  sum  of  a  finite  series. 
149.  Arithmetic  series.  —  Let  the  series  be 

(1)  S  =  a  +  (a  +  d}  +  (a  +  2d)  +  •  •  •  +  (a  +  (n  -  1)  d), 

where  d  is  the  common  difference,  a  the  first  term  and  n  the 
number  of  terms.  Write  the  same  series  in  reverse  order 

(2)  S  =  (a  +  (n  -  1)  d)  +  (a  +  (n  -  2)  d)  +  •  •  • 


Add  (1)  to  (2), 

(3)      2S  =  [a  +  (a  +  (n  -  1)  d)]  +  [a  +  (a  +  (n  -  1)  d)] 


(4)  Call  a  +  (n  -  1)  d  =  I. 

Then  on  solving  for  S,  and  using  (4),  (3)  becomes 

fK\  C  0+  f 

(5)  S=n—~  — 


ARITHMETIC  SERIES  221 

Equations  (4)  and  (5)  are  sufficient  to  solve  all  problems 
relating  to  arithmetic  series.  Of  the  five  quantities  a,  d,  n,  I,  S, 
three  must  be  known  to  find  two  for  the  two  equations  (4) 
and  (5)  are  sufficient  to  determine  two  unknowns. 

1.  How  many  terms  of  the  series  2  +  7  +  12  +  •  •  •  must 
be  taken  to  obtain  a  sum  of  at  least  100. 

He^e  a  =  2,  d  =  5,  S  =  100,  to  find  n,  and  I  if  needed. 
From  (4)  and  (5), 

<*±^  =  100, 


I  =  2  +  (n  -  1)  5. 
Eliminating  I  and  solving  for  n  gives 

1±63+ 
n  =  —  r^r  —  =  —6.2  and  +6  4. 

Since  n  must  be  a  positive  integer  and  S  must  not  be  less  than 
100,  n  must  be  taken  equal  to  7.     Hence  I  =  32. 

2.  Interpolate  25  terms  between  —16  and  36  so  as  to  form  an 
arithmetic  series. 

Here  a  =  -16,  I  =  36,  n  =  25  +  2  =  27,  to  find  d,  and  S 
if  needed.     From  (4), 

36=  -16  +  26d 
By  use  of  (5),  d  =  2. 

S  =  270. 

The  series  may  be  written  as: 

-16  -14-12-10-8-6-4-2  +  0  +  2  +  4 
+  6  +  8  +  10+12  +  14  +  16+18  +  20  +  22  +  24 
+  26  +  28  +  30  +  32  +  34  +  36. 

3.  Find  the  sum  of  13  terms  of  the  series  whose  first  three 
terms  are,  4,  3^,  2f  ,  •  •  •  . 

4.  Find  the  fifteenth  term  of  the  series, 

-  1  -  TV  +  TV  +  i  +  •  -  •  . 

5.  If  the  sum  of  an  arithmetic  series  is  500,  the  number  of 
terms,  10,  the  first  term,  0,  find  the  common  difference  and  the 
last  term. 


222  SERIES;  TRANSCENDENTAL  FUNCTIONS 

6.  A  hundred  apples  lie  on  the  ground  in  a  straight  line,  4' 
apart.     A  basket  is  4'  from  the  end  of  the  row  in  the  same  line. 
A  boy  starts  at  the  basket  and  gathers  the  apples  into  the 
basket  one  at  a  time.    How  fa*r  must  he  walk? 

7.  A  triangular  frame  is  strung  with  parallel  wires  \"  apart. 
The  first  wire  is  the  base  of  the  triangle,  the  last  \"  from  the 
vertex.     The  base  is  24"  and  the  altitude  30".     Find  the  total 
length  of  all  the  wires.     Will  the  angles  of  the  triangle  have 
any  effect  on  the  result? 

8.  The  equation  of  a  straight  line  is  y  =  2  x  -j-  4.     Ordinates 
1  unit  distance  apart  are  erected,  beginning  at  the  y-axis,  and 
ending  at  x  =  20.     Find  the  sum  of  all  these  ordinates.     Find 
the  mean  ordinate. 

9.  Find  the  sum  of  all  the  odd  integers  from  1  to  the  nth 
odd  integer  in  terms  of  n. 

10.  Find  the  sum  of  the  first  n  even  integers  in  terms  of  n. 

11.  Can  an  infinite  arithmetic  series  be  convergent ?    Why? 
149a.   Geometric  series.  —  Let  the  series  be 

(1)  S  =  a  +  ar  +  ar2  +  •  •  •  +  arn~l. 
Then 

(2)  rS  =  ar  +  or2  +  •  •  •  +  arn~l  +  arn. 

(3)  Subtracting        (r  -  1)  S  =  arn  -  a 

A  /A\  o      ar"  —  a 

and  (4)  S  =    r_1  ' 

Call  (5)  I  --=  ar"-1, 

the  last  term.    Then  substituting 

(6)  ...      S-T=T 

Equations  (5)  and  (6)  are  sufficient  to  solve  all  problems  relating 
to  geometric  series.  Any  three  of  the  five  quantities  a,  r,  I,  n 
and  S  being  given  the  other  two  can  be  found,  for  equations 
(5),  (6)  are  sufficient  to  determine  two  unknowns. 

1.  How  many  terms  of  3  +  6  +  12  +  •  •  •  must  be  taken  to 
obtain  a  sum  of  at  least  150? 

Here  a  =  3,  r  =  2,  S  =  150,  to  find  n,  I 


GEOMETRIC  SERIES  223 

By  (5),  (6), 

7.  =  3.2»- 


£i   ~~    1 

Eliminating  Z  and  solving  for  n 
2-  -51, 
log  51 


Since  n  must  be  a  positive  integer  we  must  take  n  =  6  in  order 
that  S  shall  not  be  less  than  150. 

2.  Interpolate  6  terms  between  —8  and  16  so  as  to  form  a 
geometric  series. 

Here  a  =  -8,  n  =  6  +  2  =  8,  I  =  16,  to  find  r,  and  S  if 
needed. 

By  (5),  16=  -8r7. 

Whence  r7  =  —  2 

and  7  log  r  =  log  2  (numerically), 

log  r  =  ^  =  0.0430, 

r  =  1.104  (numerically). 

But  in  this  case  r  is  negative  and  its  value  is  r  =  —1.104.  The 
series  is,  therefore, 

S  =  -8  +  8.832  -  9.751  +  10.77  -  11.89+13.9  -  14.50+16.01. 

The  result  shows  that  the  value  of  r  is  only  approximate  and 
to  get  more  accurate  results  a  more  accurate  value  of  r  must 
be  used.  It  is  noted  here  that  when  r  is  negative  every  other 
term  of  the  series  is  negative.  S  was  not  needed. 

3.  Find  the  sum  of  |  +  $  +  ^V  +  •  •  •  to  ten  terms. 

Note.  —  Find  the  tenth  term  and  then  calculate  the  sum. 
Use  equations  (5),  (6). 

4.  Find  the  first  term  of  a  series  if  the  sum  is  500,  number  of 
terms  10  and  last  term  100. 

5.  If  a  =  5.  I  =  400,  n  =  10  in  a  certain  series,  find  r  and  S. 


224  SERIES;  TRANSCENDENTAL  FUNCTIONS 

6.  If  a  +  b  +  c  +  •  •  •   is  a  geometric  series,  show  that 

-  +  r  +  -  is  a  geometric  series. 
a      b      c 

7.  If  1  +  2  +  4  +  8  +  16  +  •  •  •  form  a  geometric  series, 
continue  the  series  to  ten  terms  and  find  the  sum. 

8.  Show  that  a  +  Va6  +  6  form  a  geometric  series  if  a  and 
b  are  both  positive  or  both  negative. 

9.  Show  that  if  log  a  +  log  6  +  log  c  +  •  •  •  form  an  arith- 
metic series,  a  +  6  +  c  +  •  •  •  form  a  geometric  series. 

10.  On  a  false  balance  a  certain  object  weighs  9  Ibs.  on  one 
pan  and  16  Ibs.  on  the  other  pan.     If  x  is  the  true  weight  show 
that  9  +  x  +  12  form  a  geometric  series. 

150.  Special  case  of  geometric  series.  —  Let  n  — »  oo  and 

\r\  <  1  in 

„  _  a  —  arn 

o  —     5  • 

1  —  r 

(Note  that  this  equation  is  really  equation  (6)).     Since  |  r  \  <  1, 

lim  rn  =  0.* 

n — >°o 

Hence  the  above  equation  will  become 

a 


(1) 


1  -r 


Therefore  when  in  a  geometric  series  |  r  \  <  1  and  n  =  oo ,  the 
sum  $n  has  a  definite  limit  as  n  — >  oo .  This  fact  will  be  of  much 
use  in  later  work. 

» 

*  The  truth  of  this  is  evident  when  we  consider  any  proper  fraction, 
say  f,  and  raise  it  to  higher  and  higher  powers.  Thus 

?    -  =  f-\z    —  -  f-\3    !§  =  f-\  ^  =  (2\n 

3'    9      W  '    27      \3/  '81      \3/  '  '      '  '    3"      \3/  '  '  *  ' 

It  is  seen  that  each  successive  power  of  f  is  less  than  the  preceding 
and  that  by  taking  n  sufficiently  large  (|)n  may  be  made  as  small  as  we 
please.  To  determine  n  so  that  (f)n  <  ^,  write 

n  Gog  2  -  log  3)  <  log  A  <  -  I- 
Hence  n  (0.1761)  >  1  and  n  >  5.7      Since  n  is  an  integer,  take  n  =  6. 


HARMONIC  SERIES  225 


1.  Find  the  limit  of  the  sum  of  I+Q  +  Q+  '  *  '  +3^as 

n—  >oo. 

2.  Find  the  value  of  4.4747  ...  in  the  form  of  a  common 
fraction. 

Note.  —  Write 

g      4  |    47    ,      47 

r  100  T  10000  T 

47 
and  sum  the  geometric  series  beginning  withr^.     Add  the 

lUU 

result  to  4. 

3.  Find  the  value  of  11.911911911  ...  in  the  form  of  a 
common  fraction. 

4.  Find  the  limit  of 

i     1~\      i     T\5    i     7i      i     T\a  T    '    *    *    «•  >  "• 


i  +  /i  '  (i  +  ft)2     (i  +  /O3 
ATofe.  —  Find  the  sum  of  the  geometric  series  whose  ratio  is 

1 

1  +  /T 

5.  Find  the  sum  of  12  +  9  +  6f  +  •  •  •  to  an  infinite  num- 
ber of  terms. 

0.   If  |  r  |  <  ^,  show  that  any  term  of  a  geometric  series  is 
greater  than  the  sum  of  all  the  terms  that  follow. 

'Note.  —  Let  a  be  the  first  term  and  r  be  the  ratio,  r  <  1. 
Form  the  series  and  sum  all  terms  after  a  given  term,  compare 
the  result  with  the  given  term. 

7.   If  the  sum  of  ten  terms  of  a  geometric  series  is  244  times 
the  sum  of  the  first  five  terms,  find  the  ratio. 

151.  Harmonic  series.  —  If  the  series 

(1)  S  =  ai  +  02  +  a3  +  •  •  • 
is  a  harmonic  series,  then 

(2)  S'  =  I  +  l  +  i+  .  .  . 

<Zl         «2         #3 

is  an  arithmetic  series  and  conversely. 


226  SERIES;  TRANSCENDENTAL  FUNCTIONS 

Suppose  S  =  a  +  x-\-bisa  harmonic  series  in  which  z  is  to 
be  determined  when  a  and  b  are  given.     By  (2), 


a      z 
is  an  arithmetic  series.     Therefore 

1  _1  =  1 

x     a      b 

whence 

2ab 

x  = 


a+b 

This  value  of  x  is  called  the  harmonic  mean  of  a  and  6.    Hence 
the  series  a  +  x  +  6  above  becomes 

o 

S  =  a-\ 


. 
a  -J-  b 

In  problems  relating  to  harmonic  series  it  is  often  better  to 
take  the  reciprocals  of  the  terms  and  form  an  arithmetic  series. 
Solve  the  problem  corresponding  to  the  original  and  then  pass 
back  to  the  harmonic  series.  There  is  no  general  formula  for 
finding  the  sum  of  a  harmonic  series. 

1.  Find  the  harmonic  mean  of  3  and  7. 

2.  Form  an  equation  of  the  third  degree  whose  roots  are  in 
harmonic  series,  the  smallest  root  being  3  and  the  next  largest  4. 
See  85. 

3.  In  the  equation  of  Ex.  2,  substitute  l/y  for  x  and  deter- 
mine the  roots  of  the  resulting  equation.     Do  they  form  any 
kind  of  series  that  you  know? 

4.  Show  that  the  geometric  mean  of  two  numbers  is  the 
geometric  mean  of  their  arithmetic  mean  and  harmonic  mean. 

Note.  —  Let  a  and  6  be  the  numbers.  Form  the  means 
indicated  and  compare  as  required  in  the  problem.  Arithmetic 
mean  of  two  numbers  is  half  their  sum.  Geometric  mean  of 
two  numbers  is  the  square  root  of  their  product. 

152.  Convergence  of  series.  —  So  far  as  this  course  is 
concerned  a  series  must  be  convergent  to  be  useful  in  solving 


CONVERGENCE  OF  SERIES  227 

problems.     Only  a  few  standard  series  which  are  in  common 
use  for  studying  certain  functions  that  are  of  great  importance 
in  the  applications  of  elementary  mathematics  will  be  treated. 
Only  convergent  series  can  have  finite  and  determinate  sums. 
Consider 


The  graph  should  be  drawn  to  show  the  curve  on  both  sides  of 
x  =  1. 

A  discontinuity  occurs  at  x  =  1.    The  value  of  y  is  oo  for 
x  =  1.     By  long  division 

(1)  I-^=l  +  z  +  z2  +  .T3  + 

This  is  a  geometric  series  having  x  for  its  ratio,  and  having  an 
infinite  number  of  terms.  The  sum  of  the  series  (1)  is,  by  150, 
for  I  x  |  <1, 


1  -x 

which  is  precisely  the  function  itself.  This  shows  that  (1)  is 
an  identity  for  values  of  \x\  <  1.  When  x  =  1  both  sides  of 
(1)  become  infinite.  When  x  =  —  1  the  left  member  is  \  while 
the  right  member  is  0  or  1  according  as  n  is  even  or  odd  re- 
spectively. For  x  >  1  the  left  member  is  finite  while  the  right 
member  becomes  infinite.  The  series  on  the  right  in  (1)  can 
represent  the  function  on  the  left  only  when  |  x  |  <  1.  For  such 
values  of  z  a  finite  number  of  terms  will  give  an  approximate 
value  of  the  function. 

For  values  of  |  x  |  >  1,  a  different  series  can  be  written  which 
will  represent  the  function.     Thus 


The  last  series  is  a  geometric  series  whose  ratio  is  -  and  therefore 

x 

has  a  definite  sum  for  values  of  \x\  >  1.    A  finite  number  of 


228  SERIES;  TRANSCENDENTAL  FUNCTIONS 

terms  of  the  series  in  (2)  will  give  an  approximate  value  of  the 
function.  The  series  (2)  does  not  converge  if  |  x  \  =  1  and 
therefore  for  such  values  of  x  cannot  represent  the  function 

1 
l-x 

We  shah1  study  functions  which  can  be  represented  as  func- 
tions of  the  variable  only  in  the  form  of  series. 

153.  To  use  series  in  the  study  of  functions  it  is  necessary  to 
be  able  to  determine  in  a  given  case  whether  the  series  is  con- 
vergent. The  two  following  simple  tests  will  meet  present 
needs. 

(a)   Comparison  test.  —  Suppose 

(1)  S  =  Ui  +  W2  +  U3   '    '    '  +  Un  4-  Wn+1  +  •    •    • 

is  an  infinite  series  and  let 

(2)  S'  =  V,  +  V2  +  V3  +  •    •    •  +  Vn  +  Vn+l  +  .    .    . 

be  an  infinite  series  known  to  be  convergent.  If,  beginning  at 
any  term  of  S,  the  terms  of  S'  and  the  remaining  terms  of  S  can 
be  paired  off  in  such  a  way  that  every  term  of  S  is  less  in  abso- 
lute value  than  its  mate  in  S',  or  at  most  equal  to  it,  the  series 
S  is  convergent.  For  suppose 

|  Uk  |  <  |  Vi  |  ,      |  Uk+i  |  <  |  02  !,.-.,    |  Uk+p  |   <   |  fljH-i  !,    -    ... 

for  all  values  of  p,  where  k  is  finite.  It  is  evident  on  adding 
these  inequalities, 


Since  S'  is  convergent  it  follows  that  S  must  be  convergent. 

154.  To  employ  the  comparison  test  it  is  necessary  to  have 
several  standard  convergent  series  for  comparison  with  unknown 
series.  Let 

(1)  Sg  =  a  +  ar  +  ar2  +  •  •  •  +  arn~l  +  arn  +  •  •  •  , 

where  \r\  <  1.    This  series  was  shown  in  150  to  be  convergent. 
Let 

(2)  s,_i  +      +      +...  ++.... 


SERIES  229 


If  p  >  1,  Sp  is  convergent.    For,  write 

!  +  !<!-  J_    1__I_|_1  +  1<1- 

2?  '  SP      2P      2P~1  '    4P      5P      6P      7p      4? 

J_      1-j.l.J.,      ,+J_<8 
8p      9P  15p      8P 


Adding  these  inequalities, 


1  +-L  +  _L  + 

I  A  n 1          I         O « t         I 


This  is  a  geometric  series  whose  ratio  is  x— ?  and  is  convergent  if 


that  is,  if  p  >  1.     If  p  =  1,  ^zi  =  1  and  the  series  does  not 
converge.    Series  (2)  is  known  as  the  p-series. 

Let 
(3)  S  =  l  +  l  +  l  +  l+.  ..+!+....         ( 

Now 

l  +  i>i;   *  +  !  +  *  +  !>!  ---- 

Therefore 


By  taking  n  sufficiently  large  Sn  may  become  larger  than  any 
pre-assigned  number  M.  The  series  does  not  converge.  This 
series  is  called  the  harmonic  series. 

By  comparing  a  given  series  with  one  or  another  of  (1),  (2), 
(3),  the  convergence  or  non-convergence  of  a  number  of  series 
can  be  determined. 

1.  Prove  that  ifS  =  Wi  +  W2  +  w3+-  •  •  +un+  •  •  •  does 
not  converge,  and  if  S'  =  Vi  +  v2  +  ?'s  +  •  •  •  +  V*  +  •  •  •  is 
such  that  Vi  >  u\;  vz>  uz;  v3  >  u3;  ...  ;  vn  >  un  +  •  •  •  the 
series  S'  does  not  converge.  Consider  positive  values  only. 


230  SERIES;  TRANSCENDENTAL  FUNCTIONS 

2.   Determine  the  convergence  of: 

a.  l+     +     +  •  •  •  (use  p-series,  p  >  1). 


b.  ^ — p:  +  ~— o  +  o~~J  (compare  with  a  geometric  series). 

c.  1  H — -7=  H — -?=  +  •  •  •  (use  p-series,  p  <  1). 
155.  Ratio  test  for  convergence.  —  Let 

(1)  S  =  Ui  +  M2  +  W3  +    '    '    •    +  Un  +  Wn+i  +    •    •    •    . 

If 

Write 


lim  I-  -I   =  k  <  1,  where  k  is  a  fixed  number,  the  series 

n — >oo  \  Un  I  \ 

(1)  is  convergent. 


-  =  n    or 


-  =  r2    or    w3  = 


rn    or 


Now  if  a  common  value  can  be  assigned  to  the  r's  so  that  the 
above  inequalities  still  hold  we  may  add  and  obtain  the  result, 


+  •  •  •  +  u\rn  +  Mirn+1  +  •  •  •  • 

If,  further,  |  r  \  <  1  the  series  is  convergent.  For  the  right  side 
of  the  last  equation  then  becomes  a  geometric  series  whose  ratio 
is  less  than  one.  If  |  r  \  >  1  the  terms  of  S  increase  with  n  and 
the  series  does  not  converge.  If  |  r  \  =  1,  the  test  does  not  give 
reliable  results,  and  some  other  test  must  be  employed. 


SERIES  WITH  COMPLEX  TERMS 


231 


1.  Test  for  convergence 
1  . 


Note. — The  nth  term  is 

2.  Test  for  convergence 
1  1 


1-2-3 
1 


1-2-3 


n 


1-2-3  '   1-2-3-4-5      1.2.3.4.5.6.7  ' 
3.   Test  for  convergence  the  series  of  Ex.  2,  154. 
156.   Series  with  complex  terms.  —  The  real  parts  of  all 
terms  may  be  separated  from  the  imaginary  parts  and  arranged 
in  a  series.    The  imaginary  parts  may  also  be  arranged  in  a 
series  by  themselves.     Consider 

S  =  (fli  -f-  bii)  -{-  (#2  ~l~  bzi)  -f*  -  -  •  -f"  (fln  4~  bni)  -J-  •  •  •  . 
Call  X  =  a!  +  02  +  •  •  •  +  an  +  •  •  • 

and  F  =  &i  +  62  +  •  •  •  +  bn  +  •  -  •  . 

Since  \S\=  Vx*  +  F2,  evidently  S  is  finite  if  both  X  and  F 
are  finite.  But  X  is  finite  if  the  series  ai  +  a«  +  •  •  •  + 
On  +  •  •  •  is  convergent.  Similarly,  F  is  finite  if  the  series 
bi  +  &2  +  •  •  •  +  bn  +  -  -  •  is  convergent.  Hence  the  con- 
vergence of  the  series  of  complex  terms  follows.  The  diagram 
illustrates  the  nature  of  a  series  of  complex  terms. 

Y 


FIG.  103. 


232  SERIES;  TRANSCENDENTAL  FUNCTIONS 

n  4.  i\z     (\  4.  7*\3 
1.  Test  l  +  i+   \    o    +  T-  r^  +  '  '  '  for  convergence. 

1  •  &  L  •  4  •  a 

Note 

Q      1    ,    •  ,  1  +  2J-1  ,  -- 

~"~ 


Y  —  1  - 


1.2-3 


y  -  i    I     2 

f 


2.  For  what  values  of  x  and  y  is  the  series 
•        (g  +  *y)2 


i 
- 

convergent. 

157.  Expansion  of  functions  in  series  of  powers  of  the 
variable.  —  A  series  each  of  whose  terms  contains  a  power  of 
the  variable  is  called  a  power  series.     When  the  powers  increase 
with  n  the  series  is  called  an  ascending  power  series.     If  the 
powers  decrease  with  n  the  series  is  a  descending  power  series. 
A  series  containing  only  positive  integral  powers  of  the  variable 
is  called  an  integral  power  series. 

To  expand  a  function  it  is  first  assumed  that  the  expansion 
is  possible.  If  then  the  coefficients  of  the  terms  can  be  deter- 
mined, the  expansion  is  known  to  exist.  But  the  resulting 
series  must  be  tested  for  convergence  before  it  can  be  used  in 
calculations.  There  is  but  one  power  series  that  can  represent 
a  given  function  in  the  interval  of  convergence.  By  interval  of 
convergence  is  meant  the  aggregate  of  values  of  the  variable  for 
which  the  series  is  convergent.  If  the  interval  is  continuous 
over  a  finite  region  it  is  sufficiently  designated  by  giving  the 
bounding  values. 

158.  Consider  F  (x)  any  continuous  function  of  x  having 
derivatives  of  all  orders  which  are  continuous  in  a  given  interval. 

Let 
(1)    F(x)  =  A  +  Bx  +  Cxz  +  Dx3  +  Ex*  +  •  •  •  +  Nxn  +  -  •  • 


EXPANSION  OF  FUNCTIONS  233 

be  assumed  to  hold  for  a  certain  interval  of  x.  The  values  of 
the  coefficients  A,  B,  C,  .  .  .  are  now  to  be  determined. 
Calculate: 


(2)  F'(x) 

(3)  F"(z)=2C+2.3Z)aH-3.4#z2+  •/  •  Nn(n-l)xn~2-] 

(4)  F'"(x}*=  2-3D  +  2.3-4£k  +  •  •  • 

+  Nn\(n  -  1)  (n  -  2)  xn~3  +  • 


From  the  form  of  the  above  equations  it  is  evident  they  must 
hold  for  x  =  0,  since  A,  B,  C,  .  .  .  are  finite  constants  by 
hypothesis,  provided  x  =  0  is  in  the  region  of  convergence  of 
(1).  Therefore, 

F(Q)     =A, 
F'(0)    =B, 

F"(0)   =       2C    or     C  =  ^5, 

<u 

F'"(0)=3.2Z>    or    D  =  - 


Substituting  these  values  of  A,  B,  C,  .  .  .in  (1)  gives 

(5)     F(*)=F(0)+F/(0)x  +  ^^a?  +  ^^^+ 

_F^O)_ 
^  X 


In  this  form  F  (x)  is  said  to  be  expanded  about  the  origin. 
If  it  is  assumed  that 

(!')     F  (x)  =  A  +  B  (x  -  a)  +  C  (x  -  a)2  +  D  (x  -  a)3 
•f  •••+#(*-  a)"  +••'-, 

*  By  F'"(x)  is  meant  the  derivative  of  F"(x).     F"(x)  is  the  derivative 
of  the  second  order  of  F(x),  F'"(x)  is  the  derivative  of  the  third  order,  etc. 


234  SERIES;  TRANSCENDENTAL  FUNCTIONS 

then 

(2')  F'  (x)  =  B  +  2  C  (x  -  a)  +  3  D  (x  -  a)2  + 

(3')  F"  (x)  =  2  C  +  2  •  3  D  (x  -  a)  +  •  •  •  . 

(4')  F'"  (x)  =  2  •  3  D  +  terms  in  (x  -  a). 

In  these  equations  put  x  =  a,  and  solve  for  A,  B,  C,  .  .  .  as 
above.  Substitute  the  values  of  A,  B,  C,  ...  in  (!')  and 
obtain 

(50         F(x)  =F(a)  +  F'(a)  (x  -  a)  +  ^^  (x  -  o)« 

.  F'"  (a)  , 
+  -273-(*-«)3+  "•  ' 

In  this  expansion,  a  must  lie  in  the  region  of  convergence  of  (I/). 
In  (50  F  (x)  is  said  to  be  expanded  about  the  point  a. 
If  in  (50,  (x  +  a)  be  put  for  x  there  results 

ut    \          -pit  i  i\ 
(6) 


In  (6),  exchange  places  with  x  and  a  and  the  result  is 

"  '" 

(7) 


Each  expansion  must  be  tested  for  convergence  to  be  sure  it 
can  be  used  in  calculations. 

159.   Consider  the  binomial  series.  —  Assume 
(1)     (a+x}n=A+Bx+Cxi+Dx*+Ex*+  •  -  •  +Nxn+  •  -  -  , 
where  n  is  any  number  whatever.    Now,  by  158, 


_ 


,  l 

J.  • 

n  (n  -  1)  •  •  •  (n  -  r  +  1) 


1.2  .....  r  ,  . 

Hence  on  substituting  in  (1) 

(2)     (a  +  x)n  =  an  +  nan~lx  +  H  ^  ~  -an~*x*  +  • 

n(n-l)(n-2)  •  •  •  (n  -  r  +  1) 
+  -  1.2-3  .....  r  -a~r 


THE  BINOMIAL  SERIES 


235 


If  n  is  a  finite  positive  integer  the  series  (1)  and  (2)  are  finite. 
If  n  is  not  a  positive  integer  the  series  are  infinite  series  and  (2) 
must  be  tested  for  convergence.  We  shall  apply  the  ratio  test. 
The  (r  +  l)st  term  is 

n(n-l)  .  .  .  (n-r  + 


_ 
the  rth  term  is 


1.2-3 


u  -  n(n-l)(n-2)  .  .  .  (n-r  +  2)  .. 

1-2-3 (r  -  1) 


the  ratio 


n  —  r  +  1 


The  limit  of  this  ratio,  as  r  — >  oo ,  is Hence  the  series  is  con- 

a 

X 

vergent  if  :-   =  k  <  1,  that  is,  if  |  x  \  <  \  a  \ . 

U 

1.   Find  the  value  of  V31.    Write 
V3T  =  V36  -  5  =  (36  -  5)*  = 

fi/1  _I    A  \      6(72-5) 

\       2*36"r  /  72 

Again 


= 


(approx.). 

Note.  —  More  accurate  results  may  be  obtained  by  calcu- 
lating more  terms  of  the  series.  These  expansions  are  conver- 
gent. For  in  each  case  the  second  term  of  the  binomial  is  less 
than  the  first,  that  is,  &  <  1  and  A  <  1,  which  correspond  to 
|  x  |  <  |  a  |  in  the  formula  (2)  above. 

2.  Find  A/IO  =  v/8T"2.       4.   Find  V§3. 

3.  Find  \/18  =  v'lG  +  2.     5.   Find  ^ 


236  SERIES;  TRANSCENDENTAL  FUNCTIONS 

160.   Consider  the  expression: 
(I)     ,W 


n(n  -  1)  (n  -  2)  .  .  .  (n  -  r  +  2)  x*-1 
1-2-3  .....  (r  -  1)  n^1  ~* 

Taking  the  limit  of  each  term  as  n  —  >  oo  the  series  becomes 


Equation  (2)  defines  the  function  /  (#)  for  values  of  z  for  which 
the  series  converges.  It  is  to  be  noted  that  the  properties  of 
the  function  are  derivable  from  the  series.  By  the  ratio  test 


=  1-2-3  .....  (r  +  1) 
ur  tf 


_ 
1-2-3 


The  limit  of  this  ratio  as  ~r  —  >  oo  is  0,  41,  I,  if  x  is  finite.     Hence 
series  (2)  is  convergent  for  all  finite  values  of  x  and  /(x)  is  a 
continuous  function  for  all  finite  values  of  x. 
The  function  f(x)  will  be  called  ex  (exponential  of  x)  so  that 

(2')    f-i  +  x  +  *+*2+...  +  T-f-£—  -+.... 

When  x  =  1,  there  results 

e=l+l+$+|+...=  2.718281  .... 

Eleven  terms  are  necessary  to  obtain  this  value.     The  student 
should  make  this  calculation  in  full. 

161.  Theorem    on    logarithms.  —  Log6  y  =  log&  a  •  loga  y, 
where  a,  6.  y,  satisfy  the  definitions  of  19,  20. 

Let  logo?/  =  u     and    lo&y  =  v. 

Then  y  =  au    and    y  —  bv. 

Therefore  aw  =  6". 

a  =  6V/U. 

log&  a  =  v/u  =  log&  2//log0  y. 
That  is,  log&  y  =•  loga  y  •  log&  a. 


THE  SERIES  237 

This  is  exactly  the  theorem.  This  equation  makes  it  possible 
to  change  the  base  of  a  system  of  logarithms.  The  two  bases 
in  common  use  are  10  and  e  =  2.718281  .... 

Example.  —  Let  b  =  e,  a  =  10,  y  =  25  and  logio  25  =  1.3979. 
Tofindloge25. 

By  the  above  theorem  write, 

log™  25  =  loge  25  -  logio  e, 
whence  loge  25  =  logic  25  ,/logio  e 

or  loge  25  =  1.3979/0.4343  =  3.193. 

1.  Using  the  tables  for  finding  logarithms  to  the  base,  10,  by 
above  method  calculate  the  logarithms  of  20,  15,  35,  75,  to  the 
base  e. 

The  number,  loge  10  =  0.4343,  which  as  a  multiplier  would 
change  loge  y  to  logio  y,  is  called  the  modulus  of  the  system  of 
logarithms  to  the  base,  10. 

162.  The  series  (2'),  160,  may  be  used  to  obtain  the  deriva- 
tives of  exponential  function  and  the  logarithmic  function. 
Write 

(1)  y  =  eu, 

where  u  is  a  function  of  some  variable,  say  x.    Then 
y  -f-  Ay  =  eu+Au  =  eu  •  eAu, 
Ay  =  eu  (eAM  -  1), 


/ 

(Ax)2                          (Ax)n 

\ 

i  A  i  j_«i*  i 

2                        'l.9.             .  *j 

-*•  J 

L\y 

..N 

/ 

Ax 

/Aw      . 

Ax 

Aw                        (Aw)n-1      Aw  . 

V 

~~       1  AT               * 

A-r    '                  'l.9.            .«A«' 

j- 

= 

dx         dx' 

since  all  terms  after  the  first  vanish  with  Ax.     Now  by  (1), 
(3)  loge  y  =  w. 

By  rearranging  (2)  and  using  (1), 
(A\  du  _  ^  dx  _l 

dy  ~  eu  dx~  y 


238  SERIES;  TRANSCENDENTAL  FUNCTIONS 

Therefore 

(5)  I1*'-*- 

If  a  is  the  base  of  the  system  of  logarithms,  multiply  both  sides 
of  (4)  by  logo  e,  161,  and  there  results 

d,  1 

^Iogay  =  log06.-. 

1.  From  a  table  of  logarithms  to  the  base  e,  construct  the 
graph  of  the  equation  y  =  loge  x. 

2.  From  a  table  of  logarithms  to  the  base  10,  construct  the 
graph  of  the  equation  y  =  logio  x.     Compare  the  graphs  of  this 
exercise  with  the  graph  of  Ex.  1.     Discuss  both. 

3.  Construct  the  graphs  of  y  =  (?  and  y  =  log  x  on  the  same 
axes  to  the  same  scale.    Note  the  positions  of  the  curves. 
Each  is  the  image  of  the  other  reflected  in  a  mirror  placed  at  an 
angle  of  45°  with  the  z-axis.     Such  functions  are  inverse  to 
each  other.     See  70,  71,  graphs  of  sin  x  and  arc  sin  x. 

4.  Construct  graphs  of  y  =  xz  and  y  =  x%  on  same  axes  and 
compare  results  with  above. 

163.   It  is  often  convenient  to  use  the  logarithmic  function 
in  calculating  derivatives  of  algebraic  functions.     Consider- 

spin 

*  =        V 


(2)  log*  y  =  m  loge  x  -  n  loge  (1  -  x). 

1  .          1 


(4) 


y  dx         x         1  —  x 

dy  _  y  [m  —  (m  —  ri)  x] 
dx~          x  (1  —  x) 

(m  —  n)  xm  —  mxm~l 


1.  Find  the  derivative  of  y  =  x/(l  —  x},  by  above  method. 

2.  Find  the  derivative  of  y  =  (1  —  a1)2,  by  above  method. 

3.  Find  the  derivative  of  y  =  e?xn,  by  above  method. 


LOGARITHMIC  SERIES  239 

4.  Find  the  derivative  of  y  =  uv,  where  u  and  v  are  functions 
of  x.    Save  result  as  a  formula. 

1  +  xz 

5.  Find  the  derivative  of  y  =  r  --  •%• 

JL  •' 

6.  Calculate  the  tabular  difference  for  logarithms  of  numbers 
between  120  and  121.     (Use  Eq.  5,  162.) 

164.  It  is  now  quite  easy  to  derive  a  series  by  means  of 
which  the  logarithms  of  numbers  may  be  calculated.  Using 
Eq.  (7),  157,  write 

(1)    /(*  +  o)  =  log,  (x  +  a)  =  f  (x)  +  /'  (z)  •  a  +  f--  a2 


2-3 

The  derivatives  are 

f(x)   =  \OgeX, 


Substituting  in  (1)  gives: 


a       a2        a3 


/O\  1  /         I         \  1  I  I 

If  x  =  I  this  becomes, 

(4)  Write  also  loge  (1  —  o)  =  —a  —  -^  —  ~-  —  -j —  • 

3S        o        4 
/•<    i    _\  / 

(5)  .-.     loj 


Now  put     a  =  ~  —  —  -  in  Eq.  (5)  and  transpose 
•L  tn  ~p  1 

(6)      Iog.( 


240  SERIES;  TRANSCENDENTAL  FUNCTIONS 

This  series  converges  .rapidly  and  three  or  four  terms  are  suffi- 
cient for  calculating  logarithms  of  ordinary  numbers. 

1.  Test  series  (6)  for  convergence. 

2.  In  (6)  put  m  =  1,  2,  3,  in  succession,  using  three  terms, 
multiply  each    result  by  logio  e  =  0.4343    and    compare   the 
results  with  the  logarithms  of  2,  3,  in  the  tables. 

3.  Given  logio  10  =  1,  calculate  by  use  of  above  series  log  11. 

4.  Given  logio  3,   logio  4,  taken   from  the  tables,  calculate 
logio  13. 

e-  =  l  +  ^  +  M2  +  M3*+  ..  . 
=  !-*-+      ~4 


2    '  2-3.4 

(/y»3 
X~2T?  + 


3  '  2.3.4.5 

where  i?  =  —1.    Now  write 

yvc  _  p-ix 


(~\}  f(r 

W  J  V*J 

^  o 

and 

(2)  ft&^-^f-' 

Squaring  these  by  actual  multiplication  and  adding  gives 

[/(z)]2  +  L/i  (x)]2  =  I- 

Compare  this  result  with  equation  (1),  48.     Again  carry  out 
the  work  and  obtain  from  (1),  (2)  above, 

_  f>-i(x+y) 


Compare  this  result  with  equation  24,  53.  These  two  relations 
are  sufficient  to  show  that/  (x)  and/i  (x)  are  the  sine  and  cosine 
functions.  But  which  is  sine  and  which  is  cosine  is  not  yet 
determined.  Substituting  —x  for  x  in/  (x),  the  result  is 

/(-*)  =  -/(*)- 

*  Here  as  in  a  previous  theorem  we  have  assumed  i  to  be  a  symbol 
treated  as  a  number. 


DERIVATIVES  241 

Therefore  /  (x)  is  an  odd  function  like  the  sine  function.     By 
the  same  substitution, 

/i(-*0  =/i(«) 
and  fi  (x)  is  an  even  function  like  the  cosine  function.     It  is 

safe  to  conclude  that 

/  (x)  =  sin  x, 

/i  (x)  =  cos  x. 

The  right  members  of  (1)  and  (2)  are  known  as  Euler's  expres- 
sions for  the  sine  and  cosine  functions,  respectively. 

166.   The  derivatives  of  the  sine  and  cosine  functions  can  be 
obtained  by  use  of  (1),  (2),  165.     Write 

eiu  _  e-iu 


mi.  dy       d    .  1 

1  nen  ~r~  =  T~  sin  u  =  =-.  •  ™ 

dx      dx  2i\       dx 

_  (eiu  +  e~iu)  du 
~2 dx 
du 

=  COSM-j-- 

dx 


(1)     .-. 

Similarly 

—  sinu- 
dx 

y  = 

dy 

du 

COSU  -j-' 

dx 

giu  _j_  g—iv 

'  -iu^U 

dx 

£ 

d                     dx 

dx 

dxCOBU                  2 

_  (eiu  —  e~iu)  du 
2i        dx 

/rkN  d  .      du 

(2)  .*.     -j-  cos  u  =  —  sin  w  3— 

dx  dx 

Tt  r  x  sin  u  ,    . 

By  use  of  tan  u  = derive 

cosw 

/•0\  d  ,  „    du 

(3)  -r-  tan  u  =  sec2  u  3— 
dx  dx 


242  SERIES;  TRANSCENDENTAL  FUNCTIONS 

„    .  d  ,    du 

1.  Derive          -r-  cot  u  =  —  esc2  u  -=-• 

ax  ax 

n    TT  .  1  ,    .      a*  du 

2.  Using        sec  w  =  -  .     derive  -r-  sec  u  =  sec  w  tan  u  -r-  • 

cos  u  ax  dx 

3.  Using        cscw  =  -  —  ,    derive  -3-  cscw=  —  cscwcotw^-- 

smu  dx  dx 

4.  From  y  =  sin2  u,  find  dy/dx. 

5.  From  y  =  sin  2  u  =  2  sin  w  cos  u,  find  dy/dx  and  show 
that  cos  2  M  =  cos2  u  —  sin2  w. 

6.  From  y  =  2  sin  2  w  +  3  cos  u,  find  dy/dx. 

7.  From  ?/  =  tan  u-\-u,  find  dy/dx. 

8.  From  y  =  log  sin  w,  find  dy/dx. 

9.  From  T/  =  log  sec  u,  find  dy/dx. 

10.  From  y  =  (a  sec2  u  +  6  cos2  w)3,  find  dy/dx. 

11.  From  ?/  =  esinx,  find 


12.  From  y  =  tan2w  +  —^  find  dy/dx. 

vi 

13.  From  ?/  =  sin  u  cos  v  +  cos  u  sin  0,  find  dy/dx.     Regard 
each  term  as  a  product  of  two  functions. 

14.  From  y  =  -  ^~  ,  find  dy/dx. 

cos2w 

15.  From  y  =  sin  3  u  +  6  cos  2  u,  find  dy/dx. 

/£  3> 

16.  From  y  =  2  sin  ~  —  3  tan  ~,  find  dy/dx. 

2i  & 

167.   To  expand  sin  w  and  cos  u  in  power  series,  write  by  157  (5) 

d2    .      1 

,  -,  -r-^sinw 

/,  XJ./N       .  .-..a.  .  au          Ju  =  o    oi 

(1)  /Cw)=sinw=sinO+-r-sinw        u-\  --  ^  -  w2+  •  •  •  . 
au        Ju  =  o  ^ 

Now  -r-  sin  u  =  cos  w. 

aw 


d 
j 
aw2  aw 


sm  u  =  —  cos  u  =  —sin  u, 


a74  d  ,  , 

3—,  sin  u  =  3-  (—  cos  u)  =  sm  w. 

du*  du 


PROBLEMS  243 

Setting  u  =  0  in  each  of  these  and  substituting  the  values  in 
(1)  gives 

u3  u5 


sn  u  =  u  — 


2.3^2.3.4.5 


1.  Test  the  last  series  for  convergence.     Compare  the  last 
series  with  the  real  part  of  the  expansion  of  e**,  161. 

2.  In  a  manner  similar  to  the  above  expand  cos  u  in  a  power 
series.     Compare  the  result  with  the  imaginary  part  of  the 
expansion  for  eix,  161.     Test  the  last  series  for  convergence. 

Note  in  using  derivatives  and  series  of  the  trigonometric 
functions  the  variable  must  be  expressed  in  radians. 

TT 

3.  Using  the  series  obtained  for  sin  u,  put  u  =  ^  (?r  =  3.1416) 

and  calculate  the  sin  ^  =  sin  30°.     Compare  the  result  with 

the  value  in  a  table  of  natural  sines. 

4.  Using  the  results  obtained  above  can  you  now  prove  that 

e™  =  cosx  -f-  ismxt 

5.  Plot  graph  of  and  discuss  y  =  ex\  find  slope  at  x  =  —1, 
x  =  l. 

6.  Plot  graph  of  and  discuss  y  =  e~x\  find  slope  at  x  =  —1, 
x  =  \. 

x  =  rO  —  r  sin  0, 


7.  Plot  graph  of  and  discuss  , 

(  y  —  r  —  r  cos  6,  r  constant, 

x,  y,  the  coordinates  of  points  on  the  curve  for  values  of  6.  This 
curve  is  the  cycloid.  Read  up  in  an  encyclopedia  on  this  very 
remarkable  curve. 

8.  Find  the  slope  of  the  cycloid  at  0  =  0,  6  =  7r/4,  6  =  IT. 

dy      dy  I  dx 

Note.  —  Use  •/-  =  -£  /  -^- 

dx      d6  /  dd 

9.   Construct  and  discuss  y  =  sin  x  +  cos  x. 

10.  Construct  and  discuss  y  =  e~*  sin  x;  find  slope  at  x  =  ir/2. 

11.  Find  the  angle  at  which  y  =  sin  x  and  y  =  cos  x,  in- 
tersect. 


244  SERIES;  TRANSCENDENTAL  FUNCTIONS 

12.  Find  the  equation  of  the  tangent  to  y  =  sin  x,  at  x  = 

T/6,  X  -  7T/2. 

13.  Construct  and  discuss  r  =  ae"6  (polar  coordinates). 

14.  Construct  and  discuss  r  =  cos  3  6  (polar  coordinates). 

15.  Construct  and  discuss  y  =  cos  3  x  (rectangular  coordi- 
nates). 

167a.  In  certain  classes  of  problems  where  the  formulas  to 
be  determined  are  of  the  form  y  =  Axn,  it  is  often  easier  to  de- 
termine the  constants  from  a  graph  constructed  with  the  log- 
arithms of  the  number  pairs  instead  of  the  numbers  themselves. 
This  can  be  done  by  use  of  a  table  of  logarithms  and  ordinary 
coordinate  paper.  Of  such  frequent  use  is  this  method  in 
certain  branches  of  engineering  that  special  coordinate  paper 
is  used.  This  paper  is  divided  and  ruled  in  spaces  propor- 
tional to  the  logarithms  of  numbers  as  is  the  slide  rule.  It  is 
then  only  necessary  to  plot  with  this  paper  as  with  ordinary 
paper.  Such  paper  is  called  logarithmic  coordinate  paper. 

Suppose  we  have  the  number  pairs, 

3  10 

36  400 

0.477  1   „ 

1.556  2.602 

Locating  the  x,  y  number  pairs  directly  on  the  logarithmic 
diagram,  we  obtain  the  line  AB.  To  determine  n  measure  DC 

DC 

to  any  scale  and  4  C  to  the  same  scale.     The  ratio  j-^  =  2  is 

the  slope  of  AB  and  is  the  value  of  n.  The  value  of  A  is  the 
number  corresponding  to  1~4  or  4.  Now  substituting  these 
values  in  the  proposed  equation  y  =  Axn  there  results 

y  =  4x2 
as  the  desired  equation. 

To  see  that  these  values  are  thus  determined  write  the  pro- 
posed equation  in  the  form 

logy  =  log  A  +  nlogx. 


x 

=  1 

1. 

5 

2 

y 

=  4 

9 

16 

then 

logs 

=  0 

0. 

176 

0.301 

and 

logy 

=  0.602 

0. 

954 

1.204 

LOGARITHMIC  GRAPH 


245 


As  log  y  and  log  x  are  the  coordinates  of  points  in  this  method 
log  A  is  the  ^/-intercept  of  the  line  and  n  is  the  slope. 


I 
I 

5 

A 
3 

1  2 

0 

I» 

So 
5 

I 

t 
\ 

I 

/ 

/ 

/ 

/ 

ID 

/ 

/ 

/ 

/ 

/ 

/ 

_J_ 

/ 

/ 

A 

r* 
O 

i              g       345678  910            2        3456789  100 
Logarithmic  Scale 

FIG.  104. 

The  student  should  procure  several  sheets  of  logarithmic 
coordinate  paper  and  solve  the  above  problem  completely.  It 
should  be  noted  that  the  divisions  on  this  paper  are  exactly 
those  on  the  A  scale  of  the  Mannheim  slide  rule. 

1.  Draw  the  graph  of  the  above  problem  on  ordinary  co- 
ordinate paper  and  on  logarithmic  paper. 

Interpolate  several  points  on  each  graph  and  determine  their 
coordinates  from  each  graph.  Which  of  the  graphs  affords  the 
easiest  and  most  accurate  interpolation?  It  will  be  seen  that 
the  logarithmic  graph  has  decided  advantages. 

2.  Make  the  graph  on  logarithmic  paper  and  determine  an 
equation  of  the  form  y  =  Axn  from  the  data  below  and  also 
construct  the  ordinary  graph. 

x=l         2          3  4  5 

=  2         8         18         32         50 


246  SERIES;  TRANSCENDENTAL  FUNCTIONS 

3.  Construct  the  logarithmic  graph  and  determine  the 
equation  as  directed  in  (2)  from  the  data  below. 

x  =  10        30  40        60        80      100 

y  =  SQ        61.5        45        34        26        21.5 

Remark.  —  It  should  be  noted  that  the  logarithmic  coordinate 
paper  provides  directly  only  for  numbers  from  1  to  100.  The 
scale  is  often  5  in.  to  1  unit.  Each  of  the  two  main  divisions 
of  the  paper  provides  for  numbers  whose  logarithms  have  the 
same  characteristic.  If  numbers  from  100  to  1000  are  to  be 
represented,  the  axes,  main  division  lines  of  the  paper,  must 
be  renamed.  Ordinarily  the  left  margin  is  named  1,  the 
middle  line  10  and  the  right  margin  100.  The  axis  1  may  be 
called  10  or  100,  and  the  others  correspondingly.  This  is 
equivalent  to  changing  the  characteristic  without  disturbing 
the  mantissa  of  the  logarithms.  To  illustrate  this  the  lower 
margin  of  the  paper  with  axes  renamed  is  shown  in  the  figure 
below. 


Logarithmic 

Taper 

1 

or  10 
or  100 

10 
100 
1000 

101 
100 
1001 

FIG.  105. 

167b.  Illustrative  problem  1.  —  The"electrical  resistance  of 
a  certain  river  water  when  different  amounts  of  solids  were 
in  solution  was  found  to  be  as  follows: 

Resistance  (ohms)  y  =  1000      800      600      400      300      200. 

Solids  (parts  in  10,000,000)     x  =    215      260      340      480      615      860. 

On  the  logarithmic  paper  let  K  be  the  point  (1000,  100)  at 
first.  Lay  off  the  points  from  the  data  of  the  problem;  they 
determine  the  line  BAA1.  Drop  the  point  A1  to  A11  and  then 
move  A11  to  Am.  Draw  AmArr  parallel  to  AA1.  The  new 
names  of  the  axes  appear  underscored  on  the  diagram.  The 
point  A™  is  the  ^/-intercept  of  the  line.  For  AIUAIV  is  in  reality 
a  continuation  of  BAA1,  and  Aiy  read  to  the  new-named  axes  is 


ILLUSTRATIVE   PROBLEM   2 


247 


the  point  where  the  line  crosses  the  axis  originally  named  1. 
Since  the  graph  is  a  straight  line  on  the  logarithmic  paper  the 
equation  desired  will  be  of  the  form  y  —  Axn.  Proceeding  as 
in  the  illustrative  problem  of  34,  the  values  of  A  and  n  are 

A  =  300,000  and  n  =  -1.05. 
The  equation  is.  therefore, 

y  =  300, 

VBOOOOO  \.r 

100 —         —Sft- 


1000 


10 


Note.  —  The  sliding  of  the  graph  downward  and  to  the  left  is 
done  to  avoid  having  to  extend  the  drawing  off  the  paper  to  the 
left  and  upward.  The  slope  of  BA1  is  negative  because  the  line 
slopes  to  the  left  and  upward.  This  will  be  further  explained 
later. 

167c.  Illustrative  problem  2.  —  Let  us  determine  the  equa- 
tion of  problem  4,  36,  by  a  method  similar  to  that  of  the  last 
section.  Talcing  logarithms  of  both  sides  of  the  equation  gives 
log  p  =  log  A  +  (K  log  e)  h. 


248 


SERIES;  TRANSCENDENTAL  FUNCTIONS 


This  equation  is  of  the  first  degree  in  log  p  and  h.  This  suggests 
making  the  graph  with  a  logarithmic  scale  for  p  and  an  ordinary 
scale  for  h.  It  is  evident  the  slope  of  the  line  is  K  log  e  and  the 
intercept  on  the  p-axis  is  log  A.  The  point  A  is  30.  Measure 
AD  to  the  scale  2"  to  1  unit  and  DC  to  the  scale  2000  to  1". 

=  0.0000625.      The 


Then  the  slope  of  CA  is 
slope  is  also  K  log  e  and  log  e  =  0.4343. 

0.0000625 


Hence 


Hence  the  equation  is 


0.4343 


=  -0.00144. 


=  30e-°-00144A. 


100 


lio 


6  o 


-8000- 


2000 


4000 
Uniform  Scale 

FIG.  107. 


6000 


The  paper  ruled  to  logarithmic  scale  one  way  and  to  ordinary 
scale  the  other  way,  as  used  in  this  problem,  is  called  semi- 
logarithmic  paper. 


ILLUSTRATIVE  PROBLEM  249 

1.   Construct  on  semilogarithmic  paper  the  graph  and  deter- 
mine an  equation  of  the  form 


from  the  data  below.    A  thermometer  initially  at  19.3°  C.  is 
exposed  to  the  air  and  the  readings  taken  afterward  as  follows: 

Time  (seconds  after  starting)  t  =     0      20      40      60]     (uniform  scale) 
Temperature  C  =   19.3  14.2  10.4    7  .  6](log.  scale) 

2.  Construct  the  logarithmic  graph  and  determine  the  equa- 
tion from  the  following:   The  amount  of  water  that  will  flow 
through  a  certain  length  of  pipe  of  different  diameters  under  a 
pressure  of  50  Ibs.  per  square  inch  is  given  below: 

Diameter  in  ft.  d  =  l  1.5       2         3  46 

Quantity  water  cu.  ft.  per  sec.     q  =  4.88j  13.43  27.5  75.13  152  409 

3.  Given  the  number  pairs: 

x  =  0        1          2          3  4 

2/  =  0        ^        A-        2.7        6.4 

Draw  graph  on  logarithmic  paper  and  determine  the  equation. 

4.  The  following  data  were  observed  on  a  given  amount  of 
gas  when  no  heat  was  allowed  to  enter  or  escape: 

Pressure  p  =     20.5        25.8        54.2 
Volume    v  =       6.3          5.3          3.2 

Plot  on  logarithmic  paper  and  determine  the  relation  between 
p,  v  in  the  form  of  an  equation. 


CHAPTER  XVII 
INTEGRATION 

168.  The  derivatives  of  several  types  of  functions  have  been 
considered.  Some  of  the  applications  of  derivatives  have  been 
illustrated.  It  will  be  profitable  now  to  consider  the  inverse 
problem,  viz.:  Given  the  derivative  of  a  function,  to  determine 
the  function.  This  problem  and  the  process  involved  are 

included  under  the  name  integration.    The  sign  of  integration 

a    /* » 

is  I  .  When  this  sign  stands  before  an  expression  it  indi- 
cates that  a  new  function  is  to  be  determined  whose  derivative 
is  the  expression  under  (immediately  following)  the  sign  of 
integration.  Integration  enables  us  to  solve  a  great  number  of 
new  problems  in  science  and  geometry. 

Consider,     I  3  z2  =  y?  +  C,  where  C  is  any  constant.     To 

prove  this  equation,  take  the  derivative  of  the  right  side  and 
compare  with  the  expression  under  the  sign  of  integration. 
This  derivative  is  seen  to  be  exactly  3.r2,  which  proves  the 
truth  of  the  equation.  C  is  called  the  constant  of  integration 
and  cannot  be  determined  without  further  information.  The 
expression  x3  +  C  is  called  the  integral  of  the  expression  under 
the  sign  of  integration,  3x2.  In  what  follows,  it  is  assumed 
that  the  integrand*  is  continuous  for  values  of  the  variable 
considered. 

For  convenience  of  reference  the  derivatives  of  a  few  funda- 
mental functions  and  the  corresponding  integrals  are  given: 

*  The  integrand  is  the  expression  under  the  sign  of  integration. 

250 


DERIVATIVES  AND  INTEGRALS 


251 


4141 


14 


414  414 


14 


8 


••14  ••W  -«I4  -«I4 


252  INTEGRATION 

The  chief  difficulty  in  integration  is  to  recognize  the  type 

dit 

form  of  the  integrand  and  the  factor  to  be  used  as  -r—    To  fac- 

dx 

tor  and  arrange  an  integrand  to  fit  an  'integration  formula,  it 
is  often  necessary  to  introduce  or  take  out  a  constant  factor. 
The  selection  of  the  formula  of  integration  is  a  matter  of  judg- 
ment and  experience.  To  illustrate,  take 

fxe™  =  I  f  e3*'  •  6  z  =  £  e3*2  +  C  (formula  7). 

To  verify  the  result  find  the  derivative  of  the  integral.  This  is 
precisely  the  original  integrand  xe*x*. 

A  variable  factor  must  not  be  introduced  under  the  integral 
sign  nor  taken  from  under  the  integral  sign. 

Consider    /  (a  +  6z2)3  x.    This  may  be  written: 


8. 


.     fz*  =  ?  9.     fe2'  = 

3.  f\=1  10.  J^  +  e*'2)  =? 

4. 


13. 


7.  +  1)  (1  +  x)  =  ?        14.        3a*  =  ? 

(multiply) 

*  Each  integrand  may  be  multiplied  by  dx  before  integrating,  if  desired. 
See  176. 


DIRECTIONS  FOR  SOLVING  PROBLEMS 

15.  f^|^  =  fees-2  3  z  sin  3  z=? 
J  cos2  3  x     J 

16.  fsin3zcosz  =  ?  23.     Cf^.  =  ? 
J  J  tana; 

17.  f  (I  +  2x)  (x  +  x2)  =  ?  24.  f  csc3  a;  cot  x  =  ? 

18.  f  (1  -  cos  2  a;)  =?  25.  fx  cos  x2  =  ? 

19.  /tan  z  sec2  z  =  ?  26.  f  cos  z  •  esin  *  =  ? 

20.  f  cot2  a;  esc2  z  =  ? 

21.  /«?«-? 
.     r(l 


253 


.  J  (I  -  xrfx  = 


22 


35. 


.     f(l  +  3)  (1  -  a;)  =  ?        36.     fcos  (3  x  +  2)  =  ? 

-z2-z-l)  =?     37.     f6sin(4z2-l)z=? 
.    /  (1  -  sin  x)  cos  x  =  ?       38.    j(l  +  tan2  x)2  sec2  z  =  ? 


31. 
32. 
33. 
34. 


169.  After  some  practice  in  the  process  of  integration  some 
easy  applications  can  be  taken  up.  The  solution  of  a  problem 
will  consist  of  the  following  steps: 

1.  From  the  given  data,  formulate  the  correct  integrand. 

2.  Select  the  formula  for  integrating. 

3.  Rearrange  the  integrand,  introducing  constant  factors,  if 
necessary,  to  make  it  fit  the  formula. 

4.  Integrate  the  expression. 


254  INTEGRATION 

170.  The  function  obtained  by  integrating  will  have  a  dis- 
tinct value  for  each  value  of  the  variable  substituted  in  it.  The 
difference  between  two  values  of  the  integral  corresponding  to 
two  values  of  the  variable  is  called  a  definite  integral.  Thus  if 


be  evaluated  for  x  =  1  and  x  =  5  and  the  former  value  sub- 
tracted from  the  latter  there  will  result 

(2)  (5)4  +  C-R4+c]=624f. 


The  values  x  =  1  and  x  =  5  are  the  limits  of  integration,  the 
value  x  =  5  being  the  upper  limit  and  the  value  x  =  1  the 
lower  limit.  As  a  brief  way  to  indicate  what  was  done  in  (1) 
and  (2)  above  we  write 


/5  T5  ~[x  =  5 

*-|+C         -624*. 
0  Ji  =  l 


Now  consider  the  second  illustrative  problem  of  170.  If 
x  =  0  and  x  =  4  be  taken  as  the  limits  of  integration,  there 
will  result 


In  the  process  of  subtraction  in  each  of  the  above  cases  the 
constant  of  integration  C  was  eliminated.  This  will  always 
occur.  When  the  limits  of  integration  are  given  it  is  unneces- 
sary to  determine  C  for  it  can  be  eliminated.  An  integral  in 
which  the  undetermined  constant  of  integration  appears  is 
called  an  indefinite  integral.  All  the  integrals  of  168  are  in- 
definite integrals. 

1.  Given  the  instantaneous  speed  of  a  point  moving  in  a 
straight  line  as  a  function  of  the  time,  t,  from  some  position  of 
reference,  v  =  1  +  6  P.  Find  the  displacement  function  and 
the  space  passed  over  during  the  interval  from  the  beginning  of 
the  5th  to  the  end  of  the  12th  second. 


DEFINITE  INTEGRAL  255 

fj? 
We  have  v  =      =  1  +  6*2,  144. 


Hence 


s  =  f  (1  +  6  Z2)  =  *  +  2  Z3  +  C. 


This  Is  the  displacement  function  with  C  undetermined.     The 
space  passed  over  is  the  value  of  the  definite  integral, 

12  =  3336. 

2.  In  the  above  problem,  instead  of  the  given  data,  suppose 
that  t  =  0,  when  s  =  Q,  and  let  us  solve  the  problem  with  these 
conditions. 

We  have  from  above, 

s  =  t  +  2  f3  +  C. 

For  t  =  0,    s  =  0,    0  =  0  +  0  +  C 

and  C  =  0. 

Hence  s  =  t  +  2 13 

is  the  complete  space  function.    The  second  part  of  the  problem 
is  solved  now  as 

s  =  t  -\-  2  £*  I        =  3336 

as  above. 

3.  The  acceleration  of  a  particle  is  a  =  ktf.     Find  the  speed 
and  displacement  functions  and  determine  the  speed  and  dis- 
placement from  rest  to  t  =  t\.     See  144. 

4.  The  speed  of  emptying  a  vessel  by  a  hole  at  the  bottom 
varies  with  the  square  root  of  the  depth  of  the  hole  below  the 
surface  of  the  liquid.     How  long  will  it  take  to  empty  a  cylinder 
3'  in  diameter,   10'  high,  if  the_opening  is  1"  in  diameter? 
Given  the  speed  of  flow  v  =  V2  gh,  where  h'  is  the  depth  and 
g  =  32.16'. 

5.  The  speed  of  a  body  from  rest  is  v  =  3  tz  —  6  t  +  10. 
Find  the  acceleration  at  t  =  0  and  at  t  =  5.     Find  the  distance 
passed  over  from  rest  until  t  =  5,  t  =  10. 


256 


INTEGRATION 


171.  Area  under  a  curve.  —  The  area  under  a  curve  is  that 
portion  of  the  coordinate  plane  bounded  by  the  curve,  the 
x-axis  and  the  ordinates  at  the  ends  of  that  part  of  the  curve 
considered.  Under  this  definition  portions  of  the  area  that  may 
lie  below  the  x-axis  are  regarded  as  negative.  In  the  figure 
PQXzXi  is  the  area  under  the  curve  PABQ.  An  element  or 
increment  of  area  is  defined  as  a  narrow  strip  bounded  by  the 
arc  0,  the  ordinates  at  its  ends  and  an  element  Ax  of  the  x-axis. 


FIG.  108. 

Now  take  A  A  =  shaded  area  +  ACB 

=  y  Ax  +  9  AT/AZ, 
where  CB  =  Ay    and    0^1. 

If  </=/(*) 

is  the  equation  of  the  curve,  we  have 

AA  =  /(x)  Ax  +  0  A/(x)  •  Ax, 


dA 


since 


A/(x)  =  Ay—  »0    as    Ax—  »0. 


VOLUMES  OF  SOLIDS  OF  REVOLUTION  257 

It  follows  that 


/**,          C1* 
=         y  =         /(x) 

\J  Xi  *S  Xt 


is  the  area  under  the  curve. 

1.  Find  the  area  under  y2  =  12  x,  between  the  ordinates  for 
x  ='0and  x  =  10. 

Write 


/MO          /no     .  _      ,       v2  r  ~\x  =  10 
A=   \    y=        Vl2-x*  =  ^f          =72.5. 

«/0  «/0  i>//     Jz=0 

2.  Find  the  area  under  i/  =  x2  +  2x—  1,  between  a;  =  0, 
re  =  6. 

x      11 

3.  Find  the  area  under  o  +  f  =  1  and  the  axes. 

4.  Find  the  area  under  xy  =  12  from  x  =  1  and  a:  =  12. 

5.  Find  the  area  between  xy  —  12  and  x  -f  y  =  12. 

Note.  —  The  area  between  the  curves  is  the  difference  of  the 
areas  under  the  curves  between  the  abscissas  of  the  points  of 
intersection. 

6.  Find  the  area  under  y  =  sin  x,  from  x  =  0  to  x  =  TT. 
From  x  =  IT  to  x  =  2  ?r. 

7.  Find  the  area  of  the  triangle  whose  vertex  is  at  0,  alti- 
tude, 16,  lying  on  the  x-axis  and  base  24,  perpendicular  to  the 

x-axis. 

4 

8.  Find  area  under  y  =  -  from  x  =  1  to  x  =  10. 

*£ 

9.  Find  area  under  y  =  e?  from  x  —  —  oo  to  x  =  0. 

10.  Find  by  integration  the  area  of  a  rectangle  of  base,  6, 
and  altitude,  h. 

172.  Volumes  of  solids  of  revolution.  —  Let  y  =  f(x) 
be  a  curve.  Let  any  arc  of  the  curve,  between  two  ordinates, 
revolve  about  the  a;-axis.  The  solid  described  is  a  solid  of 
revolution.  If  an  element  of  the  curve,  AB  =  As,  revolve 
about  the  axis  a  thin  lamina  or  slice  is  generated  whose  volume  is 
A  V  =  Try*  As  +  2  irB  As  Ay 

=  TT  [/(x)]2  Ax  +  2*6  Ax  A/(x), 

where  0  =  1,  and  0—  »0  as  Ax—  >0. 


258 
Hence 

and 
Hence 


FIG.  109. 

1.   Find  the  volume  generated  by  revolving  an  arc  of  'y 
6  x  +  1,  between  z  =  0,  x  =  10,  about  the  x-axis.    Write 

_, (6  *  +  !)«. 


/MO 

=     /      T< 

t/0 


18 
=  12610  TT. 

2.  Find  the  volume  generated  by  revolving  y  =  e*  about  the 
x-axis  between  the  limits  x  =  0,  x  =  10.  Also  between  x  = 
—  oo,  x  =  0. 


THE  AVERAGE  VALUE  259 

3.  Find  the  volume  generated  by  revolving  y  =  -  about  the 

•C 

ce-axis  from  x  =  1  to  x  =  10. 

4.  Find  the  volume  generated  by  revolving  y  =  sin  x  about 
the  re-axis  from  x  =  0  to  x  =  TT. 

Note.  —  Remember  2  sin2  x  =  1  —  cos  2  re. 

5.  Find  the  volume  generated  by  revolving  y  =  ex/2  —  e~x/2 
about  the  re-axis  from  x  =  0  to  x  =  4. 

6.  Find  the  volume  bounded  by  the  two  surfaces  generated 
by  revolving  xy  =  12  and  x  +  y  =  12  about  the  re-axis. 

7.  Find  by  integration  the  volume  of  a  cylinder  of  radius,  a, 
and  altitude,  h. 

Note.  —  Consider  the  cylinder  generated  by  revolving  a 
rectangle. 

8.  Find  the  volume  generated  by  revolving  y  =  8  x  about 
the  re-axis,  from  x  =  0  to  re  =  10.     From  this  result  derive  the 
rule  for  finding  the  volume  of  a  cone  when  the  radius  and 
altitude  are  given. 

9.  Find  the  volume  generated  by  revolving  y  =  x3  about 
the  re-axis,  from  re  =  0  to  re  =  4. 

10.  Find  the  volume  generated  by  revolving  re2  +  y2  =  r2 
about  the  re-axis,  from  x  =  0  to  re  =  r.         

Note.  —  Solve  the  equation  for  y,  y  =  Vr2  —  re2. 
From  the  result  derive  the  rule  for  finding  the  volume  of  a 
sphere  of  radius,  r. 

11.  Find  the  volume  generated  by  revolving  9  re2  -f-  16  yz  = 
144  about  the  re-axis. 

12.  A  complete  meridian  of  the  earth  is  an  ellipse  whose  long 
diameter  is  7926  mi.,  and  short  diameter  7899  mi.     Write,  in 
standard  form,  the  equation  of  this  ellipse  and  by  the  method 
of  Example  11,  find  the  volume  of  the  earth  in  cu.  mi. 

173.  The  average  value  of  a  function  over  an  interval  of  the 
variable.  Let  y  =  f  (re)  be  the  function  and  Xi  <  x  <  rc2  the 
interval.  From  the  figure  it  is  evident  the  average  value  of 
the  function  is  the  average  of  the  ordinates  of  the  curve  over 
the  interval,  and  is  the  altitude  of  a  rectangle  whose  base  is 


260 


INTEGRATION 


and  whose  area  is  the  area  under  the  curve.    Therefore, 


write 


A      /w  /  A    j)fe  \ 

-  =  ya  =  —  -  or  \  -  =  2/o=  —  -  —I- 
Xz  —  xi  Xz  —  xi        \Xz-xi  X2  —  Xi/ 

This  expression  gives  the  average  value  sought.  This  is  an 
extension  of  the  idea  of  average  to  an  infinite  number  of 
values. 

Y 


FIG.  110. 


1.   Find  the  average  ordinate  of  y  =  sin  x  between  x  =  0 
and  x  =  TT.    Write 


smx 


=  -  =  0.63-K 


7T  z  =  0         7T 

Note.  —  This  example  has  important  bearing  on  the  theory 
of  dynamos  and  motors. 

2.  Find  the  average  ordinate  of  y  =  cos  x  between  x  =  0  and 
x  =  IT. 

Note.  —  Remember  areas  below  the  a>axis  are  negative. 

3.  Find  the  average  ordinate  of  y  =  cos  x  between  x  =  —  = 
and|. 


WORK  DONE  BY  A  VARIABLE  FORCE  261 

4.  Find  the  altitude  of  a  rectangle  of  base  12  that  equals  the 
area  under  y  =  —  xz  +  6  x  +  27.     Draw  figure. 

5.  Find  the  average  ordinate  of  y  =  a2  —  x2  between  x  = 
—a,  x  =  a. 

6.  Find  the  average  ordinate  of  x  +  y  =  12  between  x  =  0 
and  x  =  12. 

7.  Find  the  average  ordinate  of  y  =  ^ — ; —  between  x  =  0 

1  +  x 

andx  =  |.  Between  x  =  — 2andz  =  —  1.  Between  x  =  —  1 
and  x  =  0.  Between  x  =  —  2  and  x  =  0.  How  do  you  ex- 
plain these  results  ?  Draw  figure. 

8.  Find  the  average  ordinate  of  y  =  e~x  between   x  =  —  3 
and  x  =  3. 

9.  Find  the  average  ordinate  of  y  =  cos  x  +  sin  x  between 
x  =  ir/2  to  x  =  TT. 

10.  Find  the  average  values  of  y  =  x,  y  =  x2,  y  =  x3,  re- 
spectively, from  x  =  0  to  x  —  10. 

174.  Work*  done  by  a  variable  force.  —  Suppose  the  force 
is  expressed  as  a  function  of  some  variable,  t,  and  that  the 
displacement  is  expressed  as  a  function  of  the  same  variable. 
Thus  let  the  force  be 

/-*(*) 

and  the  displacement  in  the  direction  of  the  force 

s  =  F(f). 
Now  the  element  of  work  is 

Aw  =  As-/  +  8  A/.  As, 

where  /  is  ordinate  of  P,  CP  =  A/  and  6  =  1  (see  Fig.  llOa),  A/ 
being  positive  or  negative  according  as  the  force  is  an  increas- 
ing or  decreasing  function  with  the  displacement.  Then 

*  Work  is  defined  as  the  product  of  force  by  the  component  of  displace- 
ment in  the  direction  of  the  force. 


262 


That  is, 


INTEGRATION 
Aw  _  As    „  .   ,,As 

dw      ,ds 


$-*».*•». 


FIG.  HOo. 


Therefore 


1.   What  is  the  work  done  by  a  force  f(x)  =  10  x2  in  a  dis- 
placement <j)(x)  =  x,  from  x  =  0  to  x  =  10?    Write 

3  ~\x  =  10 

-          =  3333  units  of  wor-k. 


/MO  /no 

=   /     10  a;2- 1  =  /     10  x*  = 

t/o  Jo 


2.  It  requires  a  force  of  5  Ibs.  to  stretch  a  spring  1".  If  the 
force  required  for  any  stretch  is  proportional  to  the  stretch, 
find  the  work  done  in  stretching  the  spring  1^  ft. 

Note.  —  Write  f(x)  =  kx  and  <t>(x)  =  x. 


DIFFERENTIAL  263 

3.  The  force  of  gravity  varies  as  the  square  of  the  distance 
from  the  center  of  the  earth.     Find  the  work  required  to  lift  a 
1-lb.  mass  from  the  earth's  surface  to  a  point  500  mi.  high.     It 
is  given  that  force  is  1  Ib.  at  the  surface  and  the  radius  of  the 
earth  4000  mi. 

4.  Find  the  work  done  by  a  force  f  =  <t>(t)  =3£3  —  8£  +  l 
in  a  displacement  s  =  F(t)  =  8 1  —  6,  from  t  =  0  to  t  =  40. 

5.  A  force  varies  inversely  as  the  displacement;  when  the 
displacement  is  1  the  force  is  100.     Find  the  work  done  in  a 
displacement  of  100  from  a  displacement  of  5. 

6.  What  work  is  done  in  winding,  on  a  windlass,  a  chain  100' 
long,  weighing  2|  Ibs.  to  the  linear  foot  ? 

7.  How  much  work  is  done  by  a  force  that  can  just  roll  a 
barrel  weighing  300  Ib.  up  a  smooth  incline  of  inclination  30° 
with  the  horizontal? 

175.  So  far  we  have  regarded  the  integral  solely  as  a 
function  whose  derivative  is  given,  that  is,  as  the  inverse  of 
the  derivative.  This  viewpoint  is  fundamental  and  of  great 
value.  In  order  to  realize  more  fully  the  power  and  utility 
of  integration  as  a  mathematical  instrument  we  must  take 
another,  though  not  contradictory,  viewpoint  and  look  upon 
an  integral  as  the  sum  of  infinitesimal  elements  under  certain 
conditions.  This  idea  is  implied  in  the  second  form  of  integrals 
given  in  the  preceding  list  of  integrals. 

Let  us  now  consider  the  problem  of  171.    We  have 

(1)  AA  =  y  Ax  +  6  A?/  Ax. 

The  first  term  on  the  right,  y  Ax,  is  defined  as  the  differential 
of  A  or  "  differential  A  "  and  written  dA.  We  shall  write 
dx  (differential  x)  for  Ax  and  instead  of  the  above  equation  we 
shall  consider 

(2)  dA  =  y  dx. 

Note  that  differential  A,  (dA),  is  precisely  the  derivative 
of  A  with  respect  to  *,  (  -r- ) ,  multiplied  by  dx.  This  prin- 
ciple is  general.  It  is  now  seen  that  y  dx  is  a  rectangle  inscribed 


264  INTEGRATION 

under  the  arc  AB  and  that  the  area  under  PQ  may  be  con- 
sidered the  limit  of  the  sum  of  a  set  of  such  inscribed  rectangles 
as  their  width,  dx,  approaches  zero.  This  idea  is  similar  to 
the  one  used  in  elementary  geometry  to  show  that  a  pyramid 
is  the  limit  of  the  sum  of  a  set  of  inscribed  or  circumscribed 
prisms.  See  Wentworth,  P.  and  S.  Geom.  (rev.  ed.),  p.  312, 
or  some  other  text. 
We  may  now  write 


=   f^ydx  = 

t/Xi 


(3)     A  =   I     dA  =    I    ydx  =    I    f(x)  dx  =  F(xy)  —  F(XI), 

tS  Xi  J  X\  *J  Xi 

where  f(x)  is  the  derivative  of  F(x).  This  example  shows 
that  integrals  considered  from  the  new  viewpoint  need  no 
rules  or  processes  different  from  those  developed  for  the  inte- 
gral as  the  inverse  of  the  derivative.  The  only  difference  lies 
in  the  construction  and  interpretation  of  the  integrand. 

Example.  —  Let  the  student  revise,  according  to  the  new 
viewpoint,  each  of  the  illustrative  problems  of  172,  173,  174. 

176.  We  shall  now  apply  the  idea  of  summation  to  the 
solution  of  a  new  type  of  problem.  In  the  solution  we  shall 
need  the  idea  of  moment  81,  Ex.  5. 

Note.  —  The  centroid  of  a  system  of  parallel  forces  is  the 
point  of  application  of  their  resultant  or  equilibrant.  The 
equilibrant  is  equal  to  and  opposite  in  direction  to  the  re- 
sultant. 

The  center  of  gravity  of  a  material  body  is  the  centroid  of 
the  forces  acting  between  its  particles  and  the  earth. 

With  these  definitions  we  now  proceed  to  find  the  center 
of  gravity  of  a  thin  straight  rod  AB  of  unit  mass  per  unit 
length  and  length  1.  Let  ab  be  any  small  portion  of  AB 
of  length  dx.  Its  mass  and  the  measure  of  its  attractive 
force  may  also  be  taken  as  dx,  since  the  density  is  unity  by 
hypothesis. 

Taking  A  as  the  origin,  AB  as  the  z-axis,  I  as  the  length  of 
AB  and  x  as  the  distance  from  A  to  some  point  of  ab,  we  have 
for  the  moment  of  ab  about  A, 


CENTROID  265 

a    b  B 


x        dx 

(1)  dM  =  x  dx. 

If  we  take  the  sum  of  the  moments  of  all  such  elements  as 
dx  — >  0,  we  have 

(2)  M=  fldM  =  fl 

Jo  Jo 

The  sum  of  all  the  forces  has  the  same  measure  as  the  mass  of 
the  rod.  This  is,  of  course,  I,  which  may,  for  theoretical  con- 
siderations in  later  work,  be  looked  upon  as 

Cl          Cl          T 

(3)  m  =   I  dm  =   I  dx  =  x\  =  I. 

Jo  Jo  J 

The  distance  x  of  the  point  of  application  of  the  resultant  of 
the  forces  from  A  is  equal  to  the  moment  M  divided  by  the 
mass  (sum  of  forces)  m  or 

P 


dx 

0 


That  is,  the  center  of  gravity  of  a  thin  straight  rod  is  at  its 
center  of  length,  which  is  what  might  have  been  expected 
from  considerations  of  symmetry.  For  the  value  of  the  above 
processes  in  later  work  the  student  should  thoroughly  master 
them. 

Let  us  now  find  the  centroid  of  the  area  under  the  curve, 
y  =  f(x)  in  Fig.  108.     The  element  of  mass  (force)  is  now 

(5)  dA  =  y  dx  =  f(x)  dx. 

The  distance  of  this  element  from  OY  (conveniently  chosen  as 
the  axis  of  moments)  is  x.  We  may  now  write 

(6)  dM  =  xdA  =  xydx  =  f(x)  x  dx 

as  the  moment  of  any  element  [rectangle  under  AB].    Hence 

(7)  M  =    I    xdA  =    I    xydx  =    I    f(x)xdx  =  <f>(x2)—(l>(xi')> 

Jxi  Jxi  Jii 


266  INTEGRATION 

where  f(x)  x  is  the  derivative  of  <£  (x).    We  have  therefore 


m  A 

where  A  =  m  =  F(x?)  —  F(xJ  and/(z)  =  -j-F  (x).       See  171. 

This  is  the  abscissa  of  the  centroid.  To  find  the  ordinate 
we  may  use  the  result  of  the  first  example  and  write  for  the 
moment  of  any  element  dA  =  y  dx  about  OX 

(9)  dM=|.<M=|. 

Whe 

(10) 


2   '  2  2 

Whence 


(f(aO)2 
where  0      is  the  derivative  of  6  (x~).    Therefore 


(11) 


where  A  =  m  as  before. 

1.  Find  the  coordinates  of  the  centroid  of  the  area  under 
y2  =  12  x,  from  x  =  2  to  x  =  8. 

2.  Find  the  coordinates  of  the  centroid  of  the  area  of  the 
right  triangle  whose  vertices  are  (0,  10),  (16,  0),  (0,  0). 

Note.  —  This  is  to  be  considered  as  the  area  under  the 
hypotenuse.  Hence  find  equation  of  hypotenuse  and  proceed 
as  above. 

3.  Find  the  coordinates  of  the  centroid  of  the  area  of  the 
rectangle  whose  vertices  are  (1,  0),  (6,  0),  (1,  7),  (6,  7). 

Note.  —  Consider  this  area  as  lying  under  the  upper  side. 

4.  Find  the  distance  from  the  vertex  to  the  centroid  of  the 
triangle  whose  altitude  is  h  and  base  6. 

Note. . —  Let  the  origin  be  the  vertex  and  the  base  parallel 
to  the  y-axis.  Take  I  as  the  length  of  any  element.  Elimi- 


INTEGRATION  BY  PARTS  267 

nate  I  in  terms  of  x  by  use  of  similar  triangles  which  will  natu- 
rally occur  in  the  diagram  of  the  problem. 

5.  Find  the  distance  from  the  vertex  to  the  centroid  of  the 
cone  of  revolution  formed  by  revolving  about  OX  the  right 
triangle  whose  vertices  are  (0,  0),  (10,  0),  (10,  6). 

Note.  —  The  elements  are  now  circular  cylinders  of  altitude 
dx  and  radius  y,  where  y  is  the  ordinate  of  any  point  on  the 
hypotenuse  of  the  triangle. 

6.  Find  the  abscissa  of  the  centroid  of  the  arc  of  length  I 
and  radius  r. 

Note.  —  Draw  the  arc  so  the  origin  is  the  center  and  the 
x-axis  bisects  the  arc.  The  element  of  arc  is  da,  its  abscissa 
is  r  cos  6,  where  0  =  a/r  in  radians  measured  from  OX.  Take 

limits  from  —  ~  to  ~  or  for  6,  —  -^-  to  •=-  • 

*9  *9  s  Y  /  ¥ 

7.  Find  the  centroid  of  a  thin  rod  whose  density  varies  as 
its  distance  from  the  left  end,  the  density  being  5  at  unit  dis- 
tance, the  rod  being  16  units  long. 

Note.  —  The  element  of  mass  is  now  5  x  dx. 

8.  Solve  Ex.  4  above  if  the  density  varies  as  the  distance 
from  the  vertex  and  is  3.5  at  unit  distance. 

9.  Solve  Ex.  5,  under  same  conditions  as  Ex.  8. 

177.  Integration  by  parts.  —  One  of  the  most  useful  meth- 
ods of  integration  when  no  formula  applies  directly  is  inte- 
gration by  parts.  It  depends  on  the  possibility  of  separating 
the  given  integrand  into  two  factors  one  of  which  can  be  inte- 
grated directly.  Consider 

(1)  7!  (uv)  =  d/dx  (uv)  dx  =  udv  —  vdu; 

whence  (2)  I  udv  =  uv  —   I  v du, 

where  dv  is  the  integrable  factor  mentioned  above.     To  apply 
this  formula  consider 

r 
x  sin  x  dx. 


268  INTEGRATION 

Take  'x  =  u,  sin  x  dx  =  dv  and  substitute  in  the  formula 
I  x  sin  x  dx  =  x  ( —  cos  x)  —  I  —  cos  x  dx 

=  —x  cos  x  +  sin  x  +  C. 

In  this  example  take  sin  x  =  u  and  xdx  =  dv.  Then  we 
obtain  by  substituting  in  the  formula 

C  x2  .  Cx* 

I  x  sm  x  dx  =  -~  sin  x  —    I  -_-  cos  x  dx. 

The  last  integrand  is  more  complicated  than  the  original. 
These  results  teach  us  that  the  choice  of  factors  of  the  inte- 
grand is  a  matter  of  vital  importance.  Sometimes  only  re- 
peated trials  will  reveal  which  set  of  factors  will  lead  to  an 
integration.  Integrate  the  following: 

1.  x  log  xdx.  4.  xeaxdx. 

2.  (sec2  x  —  1)  x  dx.  5.  log  x  dx. 

3.  x  cos  xdx.  6.  xexdx. 

178.  From  171  it  is  easily  seen  that  if  the  arc  AB  is  small 
the  chord  AB  is  nearly  equal  to  it.  This  idea  will  enable 
us  to  employ  integration  to  determine  the  length  of  an  arc  of 
a  curve  when  its  equation  is  given.  For  as  in  geometry  we 
regard  the  circle  as  the  limit  of  the  sum  of  the  sides  of  an  in- 
scribed polygon  so  we  now  regard  the  curve  PQ  as  the  limit 
of  the  sum  of  such  chords  as  AB.  Hence  if  we  write 


(1)  chord  AB  =  VAC2  +  BC2  =  y  1  +  f^  AC, 

and  put  AC  =  Ax,  BC  =  Ay&ndAB  =  As  we  obtain,  noting  that 

A*  -» o  Ax      dx 

(2)  arc  PQ  = 


Sometimes  it  may  be  desirable  to  use  the  form 
(3)  >  arcP« 


INTEGRATION  269 

which  is  easily  seen  to  be  equivalent  to  the  above  formula,  the 
difference  being  that  we  now  integrate  with  respect  to  y  instead 
of  x  and  must  use  y-limits. 

1.  Find  the  length  of  an  arc  of  y*  =  x3  from  the  origin  to  the 
point  (4,  8).     From  the  given  equation 

y  =  x%> 

^/      3   t 
dx~2X  ' 

arc  =    Ida  -if  \/4  +  9xdx  =  ±  f*(4  +  9x)*dx 
Jo  Jo  Jo 


2.  Find,  the  length  of  the  catenary  y  =  &  —  e  *  from  x  =  0 
to  x  =  10. 

3.  Find  the  length  of  y  =  sin  x  from  x  =  0  to  x  =  IT. 

4.  Find  the  length  of  x$  +  y$  =  a  from  z  =  0  tojc  =  a. 

,,     c  ,     , ,,          ,  . ,  fx  =  a0  —  asinfl. 

5.  Find  the  length  of  one  arch  of  the  cycloid  < 

[y  =  a  —  a  cos  6. 

Note.  —  Remember-^  =  ,y  , ...    Use  some  formulas  from  64 
dx      dx/dd 

to  reduce  the  integrand  to  simpler  form. 

179.  Very  often  an  integrand  may  be  simplified  and  made 
to  fit  some  formula  of  integration  by  a  transformation  of  the 
variable  (see  Chap.  XIII)  or  what  is  ordinarily  called  a  sub- 

/x  dx 
— .    Let  x  =  z2.    Then 
1  +  x* 

dx  =  2  z  dz.    Substituting  in  the  given  integrand 


2  +  I 
=        -  *2  +  22  -  21og  (z  +  1)  +  C. 

Restoring  x  this  becomes 

lx$  ;  -  x  -f  2  a;*  -  \  log  (x*  +  1)  +  C. 


270  INTEGRATION 

Consider  another  example 

dx 


Let    x  =  tan  2.    Then    dx  =  sec2  z  dz    and 
Making  the  substitution  there  results 


.=   I  seczdz 
J 

/(sec  2  +  tan  z),        r  sec  2  tan  2  +  sec2  2  , 
sec  2) — (dz  =  I  —  —  dz 

(sec  z  +  tan  2)         J       sec  2  +  tan  z 

i*d(secz  +  tan  2) 


sec  2  +  tan  2 
Restoring  x  we  obtain 
dx 


=  log  (sec  2  +  tan  2)  +  C. 


Vl  +  z2 

Success  with  the  method  of  substitutions  depends  upon  wide 
experience.  The  number  of  substitution  relations  is  unlim- 
ited. The  student  should  consult  larger  texts  on  calculus  for 
further  treatment. 

1.  Integrate    I      ,  -  by  the  substitution  x  =  -• 

J  V(i  +  x2)3  y 

rx\  -  i 

2.  Integrate    I  -j dx  by  the  substitution  x  =  z4. 

J  x2  +  1 

/dx 
.  by  the  substitution  1  +  bx  =  22. 

VI  +  bx 

On  p.  276  is  a  more  extensive  table  of  integration  formulas 
than  is  given  in  168.  The  purpose  of  a  table  of  integrals  such 
as  this  is  to  save  the  time  of  the  student  and  the  practic- 
ing mathematician  after  he  has  sufficient  exercise  in  the  proc- 
ess of  integration  to  ensure  that  he  thoroughly  understands 
the  significance  of  the  formulas  that  he  is  using.  The  table 
may  be  used  in  solving  the  following  problems.  For  a  more 
extensive  table  of  integrals  the  student  is  referred  to  Hudson 
and  Lipka's  Table  of  Integrals. 


SUPPLEMENTARY  EXERCISES 


271 


ADDITIONAL  DERIVATIVES 

du 


1.  -7- (arc  sin  w)  = 


2.   -T-  (arc  tan  w)  = 


3.  -r-  (arc  sec  w)  = 


d  ,  x 

=  —  -r-  (arc  cos  w). 
da: v 


=  —  -T- (arc  tan  u). 


dw 

d   ,  N  da: 

4.  -j-  (arc  vers  u)  =      . 

da:  \/2  u  - 


tt  / 

=  —  j-  (arc  esc 
da: 


d  ,  . 

=  —  j-  (arc  covers  u). 


5.  From  vers  w  =  1  —  cos  u  find  -T-  (vers  M)  =  sin  w  -^ . 

da:  da: 

6.  From  covers  u  =  1  —  sin  u  find  3-  (covers  w)  =  —  cos  u-r- 

dx  da; 

SUPPLEMENTARY  EXERCISES 

Find  the  derivatives  of  the  following: 

* 

n      **./~9 ;   i  tan  x 

2.   x3  v  a2  —  x2  H 


1.   x*  +  e2  —  cos  x. 

3.  or*  —  x  sec  x  H — ,  • 

Vl  -x* 

5.   Log  (1  —  x2)  —  log  sin  x2. 


4.   e~zs  (1  —  ax). 

6.   3  sin  3  x  —  a  cos  nx. 


a  +  6  sin  x 
10.   sin  x  cos  3  x. 
12.   <&**  (1  -  x2). 
14.   log(l  -  3X  +  4X2). 

11.   x4  sin  5  x. 

1«      l°8  x 

13>      x2 
16    loe1"*. 

I0g  3  +  x 
17.   arc  tan  ox*. 
19.   sinx*. 

21.  arc  sec  — 

x 

18.   x4  arc  cos  x. 
20.   earcsinx. 

22.   6  sin  e~&. 

272 

Integrate  the  following: 
1.     flx*dx. 

•^  0 

dx 


•INTEGRATION 


«  vT 


7.     l    sin  x  cos  x  dx. 

Jo 

9      f**?. 

'  J     e* 

11.     j  arc  sin  x  do;. 
13.     fx2logxdx. 
15.     f  tan4  ox  dx. 


>•    fa 


tan5x 
sec3x 

dx 


dx. 


21.     f  sin2xcos2xdx. 
23.   J2i 
25.     foe 


°5dx 


/•2 

4.          x  sin  x  dx. 

Jo 


6. 

8.  |  sin2  x  dx. 

10.  fxe**dx. 

12.  J*xtan3xdx. 

14.  Jx2arcsecxdx. 

16.  Jcot3xcscxdo;. 

'   J0     1  +x2' 

IT 

/»2 

20.     I     sin3  x  cos3  x  dx. 
«/o 

o(x  —  a)3  , 


22. 
24. 
26.  / 


VlOx-25x2 


SUPPLEMENTARY  PROBLEMS 

1.  A  train  moves  out  of  station  A  on  a  straight  track.  Its  distance  s 
in  feet  from  the  station  is  given  at  any  time,  until  it  reached  its  maximum 
velocity,  by  the  equation  of  motion  s  =  80  P  +  30  I,  where  t  is  the  time  in 
minutes  since  the  train  left  the  station.  Find  how  far  the  train  travels 
during  the  first  3  minutes;  the  first  2  minutes.  Find  the  average  velocity 
of  the  train  during  the  period  of  time: 

(o)  t  =  2  until  t  =  3. 

(6)  t  =  2  until  t  =  2.5. 

(c)  t  =  2untiH  =  2.1. 

(d)  t  =  2  until  t  =  2.001. 


SUPPLEMENTARY   PROBLEMS  273 

Find  the  exact  velocity  of  the  train  when  t  =  2  and  compare  it  with  the 
average  velocities  obtained  by  the  above  arithmetical  method. 

2.  The  displacement,  s,  of  a  body  is  given  by  the  equation  s  =  5  + 
2  1 2  +  t3,  where  s  is  expressed  in  feet  and  t  in  seconds.     Find  its  average 
acceleration  for  0.001  sec.,  after  the  instant  t  =  2  and  compare  it  with  the 
exact  value  of  the  acceleration  at  the  beginning  of  the  interval. 

3.  If  s  =  i  gt2,  where  g  =  32.16  ft. /sec2  and  t  is  expressed  in  seconds, 
find  the  average  velocity  for  0.01  after  the  instant  t  =  3  and  compare  it 
with  the  exact  value  of  the  velocity  when  t  =  3. 

4.  If  a  body  moves  so  that  its  horizontal  and  its  vertical  distances  from 
the  starting  point  are,  respectively,  x  =  16  t2  and  y  =  4  t,  show  that  the 
equation  of  its  path  is  y2  =  x  and  that  its  horizontal  velocity  and  vertical 
velocity  are,  respectively,  32  T  and  4  at  the  instant  t  =  T. 

5.  If  a  ball  is  constrained  to  move  down  a  plane  inclined  at  an  angle  6 
with  the  horizontal,  the  equation  of  motion  is  s  =  -|  (g  sin  0)  t2.    Find  that 
value  of  6  that  will  give  a  maximum  velocity  for  a  given  value  of  t. 

6.  The  strength  of  a  rectangular  beam  varies  as  the  breadth  and  the 
square  of  the  depth.     Find  the  dimensions  of  the  strongest  beam  that  can 
be  cut  from  the  log  whose  diameter  is  2  a. 

7.  In  measuring  an  electric  current  by  means  of  a  tangent  galvanometer 
the  percentage  error  due  to  a  given  small  error  in  the  reading  is  proportional 

to  tan  x  + .     Show  that  this  is  a  minimum  when  x  =  45°. 

tan  x 

8.  The  force  exerted  by  a  circular  electric  current  of  radius  a  on  a  small 

magnet  whose  axis  coincides  with  the  axis  of  the  circle  varies  as ;  > 

(a2  +  z2)5 

where  x  =  distance  of  the  magnet  from  the  plane  of  the  circle.  Prove 
that  the  force  is  a  maximum  when  x  =  a/2. 

9.  Find  the  maximum  parallelepiped  that  can  be  cut  from  a  sphere  if 
one  side  of  the  base  is  twice  the  other. 

10.  Assuming  that  the  current  in  a  voltaic  cell  is  C  =  — — ^'  where  E 

T  -\-  K 

and  r  are  constants  representing  electromotive  force  and  internal  resistance 
respectively,  and  R  is  the  external  resistance,  and  that  the  power  given  out 
is  P  =  RC2;  show  that,  if  different  values  are  given  to  R,  P  will  be  a  maxi- 
mum when  R  '=  r. 

11.  A  box  is  to  be  made  from  a  square  piece  of  cardboard  a  inches  on  a 
side  by  cutting  out  squares  from  the  comers  and  turning  up  the  sides  to 
form  the  box.     Find  the  side  of  the  square  cut  out  in  order  that  the  volume 
of  the  box  may  be  a  maximum. 

12.  A  water  tank  20  ft.  high  stands  on  the  top  of  a  scaffolding  125  ft. 
high.     At  what  distance  from  the  base  of  the  scaffolding  should  one  stand 
in  order  that  the  height  of  the  tank  might  subtend  the  largest  angle  at 
the  eye? 


274  INTEGRATION 

13.  If  there  are  n  voltaic  cells  each  having  an  electromotive  force  of  e 
volts  and  internal  resistance  of  r  ohms,  and  if  x  cells  are  arranged  in  series 
and  n/x  rows  in  parallel,  the  current  that  the  battery  will  send  through  an 
external  resistance  R  is  given  by 

n  xe 

C  =  — amperes. 

—  +R 
n 

If  n  =  20,  e  =  1.9  volts,  r  =  0.2  ohm,  R  =  0.25  ohm;  how  many  cells 
must  be  in  series  to  give  the  greatest  possible  current? 

14.  If  a  body  moves  so  that  its  horizontal  and  vertical  distances  from 
a  point  are,  respectively,  x  =  10  t,  y  =  — 16  P  +  10  t,  find  its  horizontal 

16  x2 
speed  and  its  vertical  speed.     Show  that  the  path  is  y  =  +  x  and 

1UU 

that  the  slope  of  the  path  is  the  ratio  of  the  vertical  speed  to  the  horizontal 
speed. 

15.  A  point  describing  the  circle  x2  +  y2  =  25  passes  through  (3,  4) 
with  a  velocity  of  20  ft.  per  second.     Find  its  component  velocities  parallel 
to  the  axes. 

16.  A  body  moves  according  to  the  law  s  =  cos  (nt  +  e).    Show  that 
its  acceleration  is  proportional  to  the  space  through  which  it  has  moved. 

17.  Find  the  expression  for  acceleration  for  the  motion  described  by 
the  equation 

x  =  eat(ci  cos  bt  +  Cz  sin  6f). 

18.  If  a  body  is  heated  to  a  temperature  61  and  then  allowed  to  cool  by 
radiation,  its  temperature  at  the  time  t  seconds  is  given  by  the  equation 
0  =  Qtf  —  at,  where  a  is  a  constant.     Prove  that  the  rate  of  cooling  is 
proportional  to  the  temperature. 

19.  If  a  point  referred  to  rectangular  coordinates  moves  according  to 
the  law  x  =  a  cos  t  -{-  b  and  y  =  o  sin  t  +  c,  show  that  its  velocity  has  a 
constant  magnitude. 

20.  If  a  point  moves  according  to  the  law  s  =  §  gf*  +  vd*  +  SD»  find 
velocity  as  a  function  of  t,  the  acceleration  as  a  function  of  t,  and  the 
velocity  as  a  function  of  s. 

21.  For  a  beam  carrying  a  uniformly  distributed  load,  w,  per  unit 
length,  and  fixed  at  one  end, 

^  -  —  (1  -  XV 
dx*~  2EI(i 

Find  y  in  terms  of  x. 

22.  Find  expressions  for  velocity  and  distance  when  the  acceleration  is 
given  by  a  =  m  —  nfc2  cos  kt.    Determine  the  constants  of  integration, 
Ci  and  C2,  by  taking  v  =  0  and  s  =  0  when  t  =  0. 


SUPPLEMENTARY  PROBLEMS  275 

23.  In  a  chemical  reaction  of  the  first  order,  where  a  is  the  initial  con- 
centration of  a  substance  and  x  is  the  amount  of  substance  transformed  in 
a  time  1,  the  velocity  of  the  reaction  is  given  by  the  formula  dx/dt  = 
k(a  —  x).    Express  k  as  a  function  of  t  and  x,  and  x  as  a  function  of  k 
and  l. 

24.  A   point   has   an   acceleration   expressed  by   the   equation   a  = 
— no2  cos  wt,  where  r  and  w  are  constants.     Derive  expressions  for  the 
velocity  and  the  distance  traveled  if  s  =  r  and  v  =  0  when  t  =  0. 

25.  If  y  is  the  deflection  at  distance  x  from  the  fixed  end  of  uniform 
beam  of  length  I,  fixed  at  one  end  and  loaded  with  a  weight  w  at  the  other, 
then,  if  we  neglect  the  weight  of  the  beam, 

J*_fl-aa*L 

dx*      U      X)  El 

E  and  /  are  constants  depending  upon  the  material  and  shape  of  the  beam. 
It  is  known  that  the  deflection  y  and  slope  dy/dx  are  0  at  the  fixed  end 
where  x  =  0.  Find  an  expression  for  y  in  terms  of  x. 

26.  If  the  electromotive  force,  E.M.F.,  of  an  alternating  current  is 
represented  by  a  sin  curve,  any  ordinate  represents  the  E.M.F.  at  that 
point.     Show    that    Ea  =  0.637  E,   where   E  =  maximum    E.M.F.    and 
Ea  =  average  E.M.F. 

27.  Another  value  of  importance  in  the  treatment  of  alternating  cur- 
rents is  the  square  root  of  the  mean  square  of  the  ordinates,  or  the  effective 
E.M.F.    Show  that  Ee  =  0.707  E,  where  Ee  =  the  effective  E.M.F. 

28.  Air  expands  isothermally  (without  change  of  temperature)  accord- 
ing to  Boyle's  law,  pv  =  c,  where  c  is  a  constant.     The  work  done  by  such 
an  expansion  while  the  volume  changes  from  Vi  to  v2  may  be  represented  by 
the  area  under  the  curve  p  =  c/v  and  between  the  ordinates  v  =  Vi  and 
v  =  v2.    Sketch  the  curve  and  determine  this  area.     Find  the  work  done 
if  the  expansion  continues  indefinitely,  that  is,  if  v  — »  oo .     Give  a  physical 
interpretation  of  your  result. 

29.  The  equation  representing  the  adiabatic  expansion  of  a  gas  is 
pt>*  =  c,  where  c  is  a  constant.     Find  the  work  done  by  such  an  expansion 
by  finding  the  area  under  the  curve  p  —  c/vk  and  between  the  ordinates 
v  =  Vi  and  v  =  vz.     Find  the  work  done  if  v2  — >  oo .     Give  a  physical 
interpretation  of  your  result. 

Note,  —  During  the  adiabatic  expansion  of  a  gas  heat  is  not  communi- 
cated to  nor  abstracted  from  the  gas. 

30.  Derive  the  four  general  equations  of  motion,  having  given  that 
s=0  and  v  =  VD  when  t  =  0.     That  is,  find  v  as  function  of  t,  s  as  a  func- 
tion of  t,  s  as  a  function  of  t;  and  derive  the  formula  s  =  \  (vo  +  v)  t. 


276  INTEGRATION 

TABLE  OF  INTEGRALS 

C  un+l 

1.   (a)    lu»du  =  ——i  +  C.  n^ 

J  n+l 


u 


2-  <•> 

(6) 


(c)     I  enxdx  =  -enx  +  C. 
J  n 

3.    (a)  I  udv  =  uv  —  I  v  du. 

(6)  /  ze*  da;  =  ez(a;  -  1)  +  C. 

(c)  /  x^1  dz  =  e*(x*  -  2  x  +  2)  +  C. 

(d)  I  logxdx  =  x  log  #  —  x  +  C. 


4- 


t-      C       du  .    u  .    ~  u  , 

o.          /  =  arc  sin  -  +  C    or     —  arc  cos  -  +  C. 

J   Va2  —  w2  0  a 


._  u  ,  „ 

=  -arc  sec  -  +  C    or    --  arc  esc  -  +  C. 


/  0  -  -  --  - 

V  w2  —  a2      o  a  a  a 

du  1  .      u  —  a 


du  1  ,      a  + 


TABLE  OF  INTEGRALS  277 

C  du  -       1      logM~a  I   C 

J  (u  -  a)  (u  -  6)      a  -  b     gw  -  6  ^ 


8- 


±  a2 


(6)     f 
J 


. 
V(w  -  a)  (6  -  u)  6  -  a 


A          , 
V(M  —  a)  (M—  6) 

9.     /  Va2  —  w2  dw  =  £  (w  Va2  —  w2  +  a2  arc  sin  w/a)  +  C. 

10.  /Vw2±a2  dw  =  i  [w  Vw2±a2±a2  log  (t*  +  Vw2±a2)]  +  C. 

11.  /  sin  ax  da;  =  —  cos  ax  +  C. 
J  a 

12.  I  cos  ax  dx  =  -  sin  ax  +  C. 
*/  a 

/I  1 

tan  oa:  dx  =  -  log  sec  ax  +  (7  =  --  log  cos  as  +  C. 
CL  Of 

14.  I  ctn  ax  dx  =  -  log  sin  ax  +  C. 
«/  a 

C  Tsec  as  •  (tan  ax  +  sec  az)  dx 

15.  I  sec  ax  dx  =    I  -  7^  -  ;  -  r  —  -  — 
J  J  (tan  ax  +  sec  ax) 

=  -  log  (sec  ax  +  tan  ax)  +C  =  -logtan(j  +  -jr- 
a  a  \4       ^ 


1  +  sin  ax      1  —  cos  (ir/2  +  ax) 

since  [(sec  ax  +  tan  ax)  =  -    -  =  —  =  —  ,   ,'  .  -  r—  -  = 

cos  ax  sm  (r/2  +  ax) 

tart  £  (r/2  +  ox)]. 

«c      C  C  (—ctn  ox  +  csc  ox)  , 

16.     I  csc  axdx  =    I  csc  ox  •  7  -  :  -  ;  —    —  (  ax 
J  J  •  (  —  ctn  ax  +  csc  ax) 

=  -  log  (csc  ax  —  ctn  ax)  +  C  =  -  log  tan  -^  +  C. 
a  0  « 

/w      1 
sin2  M  dw  =  ^  —  T  sin  2  w  +  C. 


/M      1 
cos2  u  du  =  ^  +  T  sin  2  w  +  C. 


278  INTEGRATION 

19.     I  sec2  udu  =  tan  u  +  C. 


20 
21 


.  I  tan2  u  du  =  tan  w  —  w  +  C. 

.  /  esc2  w  dw  =  —  ctn  w  +  C. 

22.  I  sec  w  •  tan  u  du  =  sec  w  +  C. 

23.  /  esc  w  •  ctn  w  dw  =  esc  it  +  C. 

24.  /   .        =    I  esc2  u  du. 
J  sm2u      J 

25.  I  —  5—  =    I  sec2  w  du. 

J  COS2M        J 

oc      C      du  (*sec2udu      , 

26.  I  -r—         -  =    I  -  =  log  tan  u  +  C. 
J  smwcosw      J      t&nu 

27.  I  sin3  u  du  =    I  (1  —  cos2  w)  sin  w  du. 

28.  /  cos3  wdw  =    /  (1  —  sin2  w)  cos  w  dw. 

29.  /  u  sin  w  du  =  sin  w  —  u  cos  w  +  C. 

30.  I  ucosudu  =  usmu  +  cos  w  +  (7. 

/gOX 
ea*  cos  bx  dx  =    2      ,  2  (a  cos  6x  +  6  sin  6x)  +  C. 
a  ~\  o 

/eax 
sax  sin  bx  dx  =  •  „  ,   ,„  (a  sin  bx  —  b  cos  bx)  +  C. 
a2  +  62 

33.  /  arc  sin  u  du  =  u  arc  sin  u  +  Vl  —  w2  -f-  C. 

34.  /  arc  cos  u  du  =  u  arc  cos  u  —  Vl  —  uz  +  C. 

35.  /  arc  tan  u  du  =  u  arc  tan  u  —  \  log  (1  +  w2)  +  C. 

36.  /  arc  ctn  u  du  =  u  arc  ctn  u  +  $  log  (I  +  w2)  -J-  C. 


TABLE  OF  INTEGRALS 


279 


37. 

38. 
39. 
40. 
41. 
42. 
43. 
44. 
45. 
46. 


/  ware  sin  udu  =  \  [(2  uz—  1)  arc  sin  u  +  u  Vl  — w2]-f  C. 
/  ware  cos  udu  =  l  [(2  w2— 1)  arc  cos  w  -  u  Vl  —  w2]  +  C. 
/  M  arc  tan  udu  =  \  [(u2  +  1)  arc  tan  u  —  u]  +  C. 

I  w arc  ctn  udu  =  \  [(w2  +  1)  arc  ctn  w  +  u]  +  C. 

C        du  .    u    (       . 

=  =  arcversm-:  (versmw  =  1  —  cosw). 

J 


-,n  \    n 

=  log  (log  U)  +  C. 


r       dM  f 

I  -  ;  -  =     I 

J  u  •  log  w      J  log  w 

/  un  loo-  ?/  ^?y  —  ?/n+1  1      ^U  __  -  _  I  4-  (7 
J  U  +  1       (n  +  !)2/ 

f  ,du    =  =  log  (u  +  Vw2±a2). 
«/  Vw2  ±  a2 

r  du  =j- 

Ja2-w2      2a 


—du 


2aa-« 

1  ,      u  —  a 


USE   OF  TABLES 

The  use  of  tables  requires  in  general  a  knowledge  of  inter- 
polation. The  method  of  interpolation  is  illustrated  in  what 
follows.  In  making  interpolations  the  corrected  result  should 
not  contain  more  significant  figures  than  are  given  in  the 
table  which  is  used.  This  often  requires  the  "cutting  back" 
of  numbers.  For  example,  in  the  first  illustration  below,  the 
product  of  two  differences,  0.055  X  0.67  =  0.03685,  was  cut 
back  to  0.037  in  order  that  the  corrected  result  would  not 
contain  more  significant  figures  than  that  part  of  the  table 
from  which  the  corrections  were  made.  Note  that  the  last 
figure  retained  in  the  correction  was  raised  one  unit.  When 
the  part  cut  off  is  equal  to  or  greater  than  one-half  the  next 
higher  unit,  then  that  unit  is  increased  by  one  —  if  less  than 
one-half  no  change  is  made  in  that  unit. 

TABLE  I:  Powers  and  roots  of  numbers  from  i  to  100. 

Assuming  that  the  roots  of  successive  numbers  are  propor- 
tional to  their  corresponding  numbers,  we  may  find  the  root 
of  a  number  that  does  not  appear  in  the  table,  such  as  83.67, 
by  interpolation  as  follows :  From  the  table 

V84  =  9.165. 
V83  =  9.110. 

A  difference  of  1  in  the  numbers  causes  a  difference  of  0.055  in 
their  roots.  Therefore,  a  difference  of  0.67  in  the  numbers 
causes  a  difference  of  0.037  ( =  0.67  X  0.055)  in  the  roots. 

Therefore,      V83.67  =  9.110  +  0.037  =  9.147. 

The  root  of  a  decimal  fraction  such  as  the  cube  root  of  0.06831 
may  be  obtained  from  the  table  as  follows: 


USE  OF  TABLES  281 

From  the  table 

^69  =  4.102. 

^68  =  4.082. 

A  difference  of  1  in  the  numbers  causes  a  difference  of  0.02 
in  their  roots.  Therefore,  a  difference  of  0.31  in  the  numbers 
causes  a  difference  of  0.006  in  the  roots. 


Therefore,     ^68.31      -  4.082  +  0.006  =  4.088 
and  ^0.06831  =  ^  ^68.31  =  0.4088. 

In  this  illustration  notice  that  the  number  was  first  multi- 
plied by  1000  and  the  cube  root  of  this  new  number  was  found 
from  the  table.  This  root  when  divided  by  10  gave  the  root 
sought. 

To  find  the  square  root  of  the  decimal  fraction,  0.06831, 
multiply  the  number  by  100.     Then  find  the  root  of  this  num- 
ber (6.831)  and  divide  this  root   (2.613)  by  10  and  obtain 
0.2613  as  the  square  root  of  0.06831. 
TABLE  II:  How  to  find  the  logarithm  of  a  number. 

Find  log  27.4.     The  characteristic  by  rule  is  1. 

To  obtain  the  mantissa  from  the  table,  look  for  27  in  the 
column  headed  N.  Opposite  27  and  in  the  column  headed  by 
4  find  the  number  4378,  which  is  the  mantissa  with  decimal 
point  omitted.  Therefore,  log  27.4  =  1.4378. 

Find  log  274.3.     The  mantissa  of  this  logarithm  is  not  given 
directly  in  the  table.     If  we  assume  that  a  small  change  in 
the  number  causes  a  proportional  change  in  the  logarithm, 
then  we  may  proceed  by  interpolation  as  follows: 
mantissa  of  log  275  is  0.4393. 
mantissa  of  log  274  is  0.4378. 

We  observe  that  a  difference  of  1  in  the  number  makes  a 
difference  of  0.0015  in  the  logarithm,  or  a  difference  of  0.3  in 
the  number  makes  a  difference  of  0.0015  X  0.3  =  0.00045,  or 
cutting  the  number  back  one  place,  our  correction  is  0.0005. 
The  logarithm  of  274.3  =  2.4378  +  0.0005  =  2.4383.  Time 
will  be  saved  in  interpolating  by  considering  the  mantissas  for 


282  USE  OF  TABLES 

the  moment  as  whole  numbers.    Thus,   instead  of  writing 
0.0015,  write  15,  and  so  forth. 

How  to  find  the  number  corresponding  to  a  given  loga- 
rithm. —  Given  log  N  =  2.4383.  To  find  the  number  N. 
Since  this  problem  is  the  converse  of  the  preceding  one,  we 
may  trace  that  problem  back.  We  cannot  find  the  mantissa 
0.4383  in  the  table,  but  we  find  0.4393  and  0.4378  and  so  forth. 

Mechanical  interpolations.  —  In  order  to  facilitate  the  com- 
putation, the  tabular  difference  and  the  proportional  part  for 
the  fourth  figure  of  the  natural  numbers  is  given  at  the  bot- 
tom of  the  page.  The  student  is  advised  not  to  use  this  part 
of  the  table  until  he  has  learned  to  interpolate  mentally  with 
speed  and  accuracy.  In  scientific  work  one  is  called  upon  to 
use  many  different  tables  in  which  tabular  differences  and  pro- 
portional parts  are  not  given.  For  this  reason,  the  student 
should  learn  to  be  independent  of  these  aids. 

In  the  above  problems  beginning  with  the  tabular  difference 
4393  -  4378  =  15,  look  at  bottom  of  page  under  "Tab.  Difif." 
for  15  which  is  found  on  the  third  page  of  the  tables.  Opposite 
15  and  in  column  headed  3  find  4.5.  Add  this  correction, 
after  cutting  it  back  according  to  rule,  to  the  mantissa  4378 
and  obtain  4383,  the  same  mantissa  as  above  with  decimal 
point  omitted. 

If  log  N  =  2.4383,  find  the  number  N,  proceeding  as  follows: 
Find  mantissa  in  the  table  nearest  less  than  4383.  It  is  4378, 
which  is  in  column  headed  4  and  opposite  the  number  27  in 
the  first  column.  Hence  0.4378  is  the  mantissa  of  the  loga- 
rithm of  274. 

Now  4383-4378=    5. 

4393  -  4378  =  15. 

At  bottom  of  page  under  "Tab.  Diff."  opposite  15,  find  the 
number  nearest  5.  It  is  4.5  in  column  headed  3.  Therefore, 
3  is  the  next  figure  (fourth)  of  the  number  sought.  The  deci- 
mal point,  by  rule  for  characteristic,  should  be  placed  so  that 
the  integral  part  of  the  number  will  have  three  digits.  There- 
fore, the  number  is  274.3. 


USE  OF  TABLES  283 

Use  of  Tables  of  Trigonometric  Functions 

In  Table  III  are  given  the  natural  and  logarithmic  functions 
of  angles  for  every  10  minutes  in  the  quadrant.  The  angles 
less  than  45°  are  found  in  the  left-hand  column.  The  func- 
tions are  given  in  same  line  with  the  angle.  Angles  from 
45°  to  90°  are  found  in  the  right-hand  column.  It  should  be 
noted  that  when  angles  are  read  on  left,  function  names  are 
to  be  read  at  top  of  page  and  when  angles  are  read  in  right 
column,  function  names  are  to  be  read  at  bottom  of  page. 

To  find  the  logarithmic  sine  of  41°  20.'  —  Since  this  angle 
is  less  than  45°  read  the  angle  in  left  column.  On  the  line  of 
41°  20'  in  column  headed  log  sine  find 

log  sin  41°  20'  =  9.8198. 

The  table  is  so  constructed  that  the  logarithmic  functions  are 
10  larger  than  their  actual  values.  This  will  be  understood  if 
the  logarithm  of  the  natural  sine  of  41°  20'  is  found  and  its 
logarithm  found  in  Table  II.  The  adding  of  10  to  the  loga- 
rithmic functions  is  a  matter  of  facility  in  calculating  with  them. 
To  find  an  angle  corresponding  to  a  given  function  we 
proceed  in  a  manner  similar  to  finding  a  number  when  its 
logarithm  is  given. 


284 


TABLES 


I.  POWERS  AND  ROOTS  OF   NUMBERS  FROM  1  TO  100 


No. 

Square. 

Cube. 

Square  root. 

Cube  root. 

1 

1 

1 

1.000 

1.000 

2 

4 

8 

1.414 

1.260 

3 

9 

27 

1.732 

1.442 

4 

16 

64 

2.000 

1.587 

5 

25 

125 

2.236 

1.710 

6 

36 

216 

2.450 

1.817 

7 

49 

343 

2.646 

1.913 

8 

64 

512 

2.828 

2.000 

9 

81 

729 

3.000 

2.080 

10 

100 

1000 

3.162 

2.154 

11 

121 

1331 

3.317 

2.224 

12 

144 

1728 

3.464 

2.289 

13 

169 

2197 

3.606 

2.351 

14 

196 

2744 

3.742 

2.410 

15 

225 

3375 

3.873 

2.466 

16 

256 

4096 

4.000 

2.520 

17 

289 

4913 

4.123 

2.571 

18 

324 

5832 

4.243' 

2.621 

19 

361 

6859 

4.359 

2.668 

20 

400 

8000 

4.472 

2.714 

21 

441 

9261 

4.583 

2.759 

22 

484 

10648 

4.690 

2.802 

23 

529 

12167 

4.796 

2.844 

24 

576 

13824 

4.899 

2.885 

25 

625 

15625 

5.000 

2.924 

26 

676 

17576 

5.099 

2.963 

27 

729 

19683 

5.196 

3.000 

28 

784 

21952 

5.292 

3.037 

t  29 

841 

24389 

5.385 

3.072 

30 

900 

27000 

5.477 

3.107 

31 

961 

29791 

5.568 

3.141 

32 

1024 

32768 

5.657 

3.175 

33 

1089 

35937 

5.745 

3.208 

34 

1156 

39304 

5.831 

3.240 

35 

1225 

42875 

5.916 

3.271 

36 

1296 

46656 

6.000 

3.302 

37 

1369 

50653 

6.083 

3.332 

38 

1444 

54872 

6.164 

3.362 

39 

1521 

59319 

6.245 

3.391 

40 

1600 

64000 

6.325 

3.420 

41 

1681 

68921 

6.403 

3.448 

42 

1764 

74088 

6.481 

3.476 

43 

1849 

79507 

6.557 

3.503 

44 

1936 

85184 

6.633 

3.530 

45 

2025 

91125 

6.708 

3.557 

46 

2116 

97336 

6.782 

3.583 

47 

2209 

103823 

6.856 

3.609 

48 

2304 

110592 

6.928 

3.634 

49 

2401 

117649 

7.000 

3.659 

50 

2500 

125000 

7.071 

3.684 

TABLES 


285 


I.   POWERS  AND  ROOTS  OF  NUMBERS  FROM  1  TO  100.  (Cont'd) 


No. 

Square. 

Cube. 

Square  root. 

Cube  root. 

51 

2601 

132651 

7.141 

3.708 

52 

2704 

140608 

7.211 

3.733 

53 

2809 

148877 

7.280 

3.756 

54 

2916 

157464 

7.349 

3.780 

55 

3025 

166375 

7.416 

3.803 

56 

3136 

175616 

7.483 

3.826 

57 

3249 

185193 

7.550 

3.849 

58 

3364 

195112 

7.616 

3.871 

59 

3481 

205379 

7.681 

3  893 

60 

3600 

216000 

7.746 

3.915 

61 

3721 

226981 

7.810 

3.937 

62 

3844 

238328 

7.874 

3.958 

63 

3969 

250047 

7.937 

3.979 

64 

4096 

262144 

8.000 

4.000 

65 

4225 

274625 

8.062 

4.021 

66 

4356 

287496 

8.124 

4.041 

67 

4489 

300763 

8.185 

4.062 

68 

4624 

314432 

8.246 

4.082 

69 

4761 

328509 

8.306 

4.102 

70 

4900 

343000 

8.367 

4.121 

71 

5041 

357911 

8.426 

4.141 

72 

5184 

373248 

8.485 

4.160 

73 

5329 

389017 

8.544 

4.179 

74 

5476 

405224 

8.602 

4.198 

75 

5625 

421875 

8.660 

4.217 

76 

5776 

438976 

8.718 

4.236 

77 

5929 

456533 

8.775 

4.254 

78 

6084 

474552 

8.832 

4.273 

79 

6241 

493039 

8.888 

4.291 

80 

6400 

512000 

8.944 

4.309 

81 

6561 

531441 

9.000 

4.327 

82 

6724 

551368 

9.055 

4.345 

83 

6889 

571787 

9.110 

4.362 

84 

7056 

592704 

9.165 

4.380 

85 

7225 

614125 

9.220 

4.397 

86 

7396 

636056 

9.274 

4.414 

87 

7569 

658503 

9.327 

4.431 

88 

7744 

681472 

9.381 

4.448 

89 

7921 

704969 

9.434 

4.465 

90 

8100 

729000 

9.487 

4.481 

91 

8281 

753571 

9.539 

4.498 

92 

8464 

778688 

9.592 

4.514 

93 

8649 

804357 

9.644 

4.531 

94 

8836 

830584 

9.695 

4.547 

96 

9025 

857375 

9.747 

4.563 

96 

9216 

884736 

9.798 

4.579 

97 

9409 

912673 

9.849 

4.595 

98 

9604 

941192 

9.900 

4.610 

99 

9801 

970299 

9.950 

4.626 

100 

10000 

1000000 

10.000 

4.642 

286 


TABLES 


II.    LOGARITHMS 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

Io~ 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

Tab.  diff. 

Extra  digit. 

1 

2 

3 

4 

5 

6 

7 

8 

9 

43 

4.3 

8.6 

12.9 

17.2 

21.5 

25.8 

30.1 

34.4 

38.7 

42 

4.2 

8.4 

12.6 

16.8 

21.0 

25.2 

29.4 

33.6 

37.8 

41 

4.1 

8.2 

12.3 

16.4 

20.5 

24.6 

28.7 

32.8 

36.9 

40 

4.0 

8.0 

12.0 

16.0 

20.0 

24.0 

28.0 

32.0 

36.0 

39 

3.9 

7.8 

11.7 

15.6 

19.5 

23.4 

27.3 

31.2 

35.1 

38 

3.8 

7.6 

11.4 

15.2 

19.0 

22.8 

26.6 

30.4 

34.2 

37 

3.7 

7.4 

11.1 

14.8 

18.5 

22.2 

25.9 

29.6 

33.3 

36 

3.6 

7.2 

10.8 

14.4 

18.0 

27.6 

25.2 

28.8 

32.4 

35 

3.5 

7.0 

10.5 

14.0 

17.5 

21.0 

24.5 

28.0 

31.5 

34 

3.4 

6.8 

10.2 

13.6 

17.0 

20.4 

23.8 

27.2 

30.6 

33 

3.3 

6.6 

9.9 

13.2 

16.5 

19.8 

23.1 

26.4 

29.7 

32 

3.2 

6.4 

9.6 

12.8 

16.0 

19.2 

22.4 

25.6 

28.8 

31 

3.1 

6.2 

9.3 

12.4 

15.5 

18.6 

21.7 

24.8 

27.9 

30 

3.0 

6.0 

9.0 

12.0 

15.0 

18.0 

21.0 

24.0 

27.0 

TABLES 


287 


II.   LOGARITHMS    (Continued) 


No. 

0 

l 

2 

3 

4 

5 

6 

7 

8 

9 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

61 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241' 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

Tab.  diff. 

Extra  digit. 

1 

2 

3 

4 

5 

6 

7 

8 

9 

29 

2.9 

5.8 

8.7 

11.6 

14.5 

17.4 

20.3 

23.2 

26.1 

28 

2.8 

5.6 

8.4 

11.2 

14.0 

16.8 

19.6 

22.4 

25.2 

27 

2.7 

5.4 

8.1 

10.8 

13.5 

16.2 

18.9 

21.6 

24.3 

26 

2.6 

5.2 

7.8 

10.4 

13.0 

15.6 

18.2 

20.8 

33.4 

25 

2.5 

5.0 

7.5 

10.0 

12.5 

15.0 

17.5 

20.0 

22.5 

24 

2.4 

4.8 

7.2 

9.6 

12.0 

14.4 

16.8 

19.2 

21.6 

23 

2.3 

4.6 

6.9 

9.2 

11.5 

13.8 

16.1 

18.4 

20.7 

22 

2.2 

4.4 

6.6 

8.8 

11.0 

13.2 

15.4 

17.6 

19.8 

21 

2.1 

4.2 

6.3 

8.4 

10.5 

12.6 

14.7 

16.8 

18.9 

20 

2.0 

4.0 

6.0 

8.0 

10.0 

12.0 

14.0 

16.0 

18.0 

19 

1.9 

3.8 

5.7 

7.6 

9.5 

11.4 

13.3 

15.2 

17.1 

18 

1.8 

3.6 

5.4 

7.2 

9.0 

10.8 

12.6 

14.4 

16.2 

17 

1.7 

3.4 

5.1 

6.8 

8.5 

10.2 

11.9 

13.6 

15.3 

16 

1.6 

3.2 

4.8 

6.4 

8.0 

9.6 

11.2 

12.8 

14.4 

288 


TABLES 


II.   LOGARITHMS    (Continued) 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

Tab.  Diff. 

Extra  digit. 

1 

2 

3 

4 

5 

6 

7 

8 

9 

15 

1.5 

3.0 

4.5 

6.0 

7.5 

9.0 

10.5 

12.0 

13.5 

14 

1.4 

2.8 

4.2 

5.6 

7.0 

8.4 

9.8 

11.2 

12.6 

13 

1.3 

2.6 

3.9 

5.2 

6.5 

7.8 

9.1 

10.4 

11.7 

12 

1.2 

2.4 

3.6 

4.8 

6.0 

7.2 

8.4 

9.6 

10.8 

11 

1.1 

2.2 

3.3 

4.4 

5.5 

6.6 

7.7 

8.8 

9.9 

10 

1.0 

2.0 

3.0 

4.0 

5.0 

6.0 

7.0 

8.0 

9.0 

9 

0.9 

1.8 

2.7 

3.6 

4.5 

5.4 

6.3 

7.2 

8.1 

8 

0.8 

1.6 

2.4 

3.2 

4.0 

4.8 

5.6 

6.4 

7.2 

7 

0.7 

1.4 

2.1 

2.8 

3.5 

4.2 

4.9 

5.6 

6.3 

6 

0.6 

1.2 

1.8 

2.4 

3.0 

3.6 

4.2 

4.8 

5.4 

5 

0.5 

1.0 

1.5 

2.0 

2.5 

3.0 

3.5 

4.0 

4.5 

4 

0.4 

0.8 

1.2 

1.6 

2.0 

2.4 

2.8 

3.2 

3.6 

TABLES 


2S9 


III.    NATURAL  AND  LOGARITHMIC  FUNCTIONS 


Sine. 

Cosine. 

Tangent. 

Cotangent. 

Angle. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

0° 

0.0000 

—  00 

1.0000 

0.0000 

0.0000 

—  00 

00 

00 

90° 

10' 

.0029 

7.4637 

1.0000 

.0000 

.0029 

7.4637 

343.8 

12.5363 

50' 

20' 

.0058 

.7648 

1.0000 

.0000 

.0058 

.7648 

171.9 

.2352 

40' 

30' 

.0087 

.9408 

1.0000 

.0000 

.0087 

.9409 

114.6 

.0591 

30' 

40' 

.0116 

8.0658 

0.9999 

.0000 

.0116 

8.0658 

85.94 

11.9342 

20' 

50' 

.0145 

.1627 

.9999 

.0000 

.0145 

.1627 

68.75 

.8373 

10' 

1° 

.0175 

.2419 

.9998 

9.9999 

.0175 

.2419 

57.29 

.7581 

89° 

10' 

.0204 

.3088 

.9998 

.9999 

.0204 

.3089 

49.10 

.6911 

50' 

20' 

.0233 

.3668 

.9997 

.9999 

.0233 

.3669 

42.96 

.6331 

40' 

30' 

.0262 

.4179 

.9997 

.9999 

.0262 

.4181 

38.19 

.5819 

30' 

40' 

.0291 

.4637 

.9996 

.9998 

.0291 

.4638 

34.37 

.5362 

20' 

50' 

.0320 

.5050 

.9995 

.9998 

.0320 

.5053 

31.24 

.4947 

10' 

2° 

.0349 

.5428 

.9994 

.9997 

.0349 

.5431 

28.64 

.4569 

88° 

10' 

.0378 

.5776 

.9993 

.9997 

.0378 

.5779 

26.43 

.4221 

50' 

1  20' 

.0407 

.6097 

.9992 

.9996 

.0407 

.6101 

24.54 

.3899 

40'  ft 

•§  30' 

.0436 

.6397 

.9990 

.9996 

.0437 

.6401 

22.90 

.3599 

30' 

1  40' 

.0465 

.6677 

.9989 

.9995 

.0466 

.6682 

21.47 

.3318 

20' 

«§  50' 

.0494 

.6940 

.9988 

.9995 

.0495 

.6945 

20.21 

.3055 

10'  * 

3° 

.0523 

.7188 

.9986 

.9994 

.0524 

.7194 

19.08 

.2806 

87° 

10' 

.0552 

.7423 

.9985 

.9993 

.0553 

.7429 

18.08 

.2571 

50' 

20' 

.0581 

.7645 

.9983 

.9993 

.0582 

.7652 

17.17 

.2348 

40' 

30' 

.0610 

.7857 

.9981 

.9992 

.0612 

.7865 

16.35 

.2135 

30' 

40' 

.0640 

.8059 

.9980 

.9991 

.0641 

.8067 

15.61 

.1933 

20' 

50' 

.0669 

.8251 

.9978 

.9990 

.0670 

.8261 

14.92 

.1739 

10' 

4° 

.0698 

.8436 

.9976 

.9989 

.0699 

.8446 

14.30 

.1554 

86° 

10' 

.0727 

.8613 

.9974 

.9989 

.0729 

.8624 

13.73 

.1376 

50' 

20' 

.0756 

.8783 

.9971 

.9988 

.0758 

.8795 

13.20 

.1205 

40' 

30' 

.0785 

.8946 

.9969 

.9987 

.0787 

.8960 

12.71 

.1040 

30' 

40' 

.0814 

.9104 

.9967 

.9986 

.0816 

.9118 

12.25 

.0882 

20' 

50' 

.0843 

.9256 

.9964 

.9985 

.0846 

.9272 

11.83 

.0728 

10' 

5° 

.0872 

.9403 

.9962 

.9983 

.0875 

.9420 

11.43 

.0580 

85° 

10' 

.0901 

.9545 

.9959 

.9982 

.0904 

.9563 

11.06 

.0437 

50' 

20' 

.0929 

.9682 

.9957 

.9981 

.0934 

.9701 

10.71 

.0299 

40' 

30' 

.0958 

.9816 

.9954 

.9980 

.0963 

.9836 

10.39 

.0164 

30' 

40' 

.0987 

.9945 

.9951 

.9979 

.0992 

.9966 

10.08 

.0034 

20' 

50' 

.1016 

9.0070 

.9948 

.9977 

.1022 

9.0093 

9.788 

10.9907 

10' 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

A  n~1A 

Cosine. 

Sine. 

Cotangent. 

Tangent. 

Ann  !e. 

For  angles  over  45°  use  right  column  and  take  names  of  functions  at 
bottom  of  page. 


290 


TABLES 


III.    NATURAL  AND   LOGARITHMIC   FUNCTIONS    (Continued) 


Sine. 

Cosine. 

Tangent. 

Cotangent. 

Angle. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

6° 

0.1045 

9.0192 

0.9945 

9.9976 

0.1051 

9.0126 

9.514 

10.9784 

84° 

10' 

.1074 

.0311 

.9942 

.9975 

.1080 

.0336 

9.255 

.9664 

50' 

20' 

.1103 

.0426 

.9939 

.9973 

.1110 

.0453 

9.010 

.9547 

40' 

30' 

.1132 

.0539 

.9936 

.9972 

.1139 

.0567 

8.777 

.9433 

30' 

40' 

.1161 

.0648 

.9932 

.9971 

.1169 

.0678 

8.556 

.9322 

20' 

50' 

.1190 

.0755 

.9929 

.9969 

.1198 

.0786 

8.345 

.9214 

10' 

7° 

.1219 

.0859 

.9925 

.9968 

.1228 

.0891 

8.144 

.9109 

83° 

10' 

.1248 

.0961 

.9922 

.9966 

.1257 

.0995 

7.953 

.9005 

50' 

20' 

.1276 

.1060 

.9918 

.9964 

.1287 

.1096 

7.770 

.8904 

40' 

30' 

.1305 

.1157 

.9914 

.9963 

.1317 

.1194 

7.596 

.8806 

30' 

40' 

.1334 

.1252 

.9911 

.9961 

.1346 

.1291 

7.429 

.8709 

20' 

50' 

.1363 

.1345 

.9907 

.9959 

.1376 

.1385 

7.269 

.8615 

10' 

8° 

.1392 

.1436 

.9903 

.9958 

.1405 

.1478 

7.115 

.8522 

82° 

10' 

.1421 

.1525 

.9899 

.9956 

.1435 

.1569 

6.968 

.8431 

50' 

I  20' 

.1449 

.1612 

.9894 

.9954 

.1465 

.1658 

6.827 

.8342 

40'  §• 

-8  30' 

.1478 

.1697 

.9890 

.9952 

.1495 

.1745 

6.691 

.8255 

30'  -g 

•8  40' 

.1507 

.1781 

.9886 

.9950 

.1524 

.1831 

6.561 

.8169 

20'  £ 

«S  50' 

.1536 

.1863 

.9881 

.9948 

.1554 

.1915 

6.435 

.8085 

10' 

9° 

.1564 

.1943 

.9877 

.9946 

.1584 

.1997 

6.314 

.8003 

81° 

10' 

.1593 

.2022 

.9872 

.9944 

.1614 

.2078 

6.197 

.7922 

50' 

20' 

.1622 

.2100 

.9868 

.9942 

.1644 

.2158 

6.084 

.7842 

40' 

30' 

.1650 

.2176 

.9863 

.9940 

.1673 

.2236 

5.976 

.7764 

30' 

40' 

.1679 

.2251 

.9858 

.9938 

.1703 

.2313 

5.871 

.7687 

20' 

50' 

.1708 

.2324 

.9853 

.9936 

.1733 

.2389 

5.769 

.7611 

10' 

10° 

.1736 

.2397 

.9848 

.9934 

.1763 

.2463 

5.671 

.7537 

80° 

10' 

.1765 

.2468 

.9843 

.9931 

.1793 

.2536 

5.576 

.7464 

50' 

20' 

.1794 

.2538 

.9838 

.9929 

.1823 

.2609 

5.485 

.7391 

40' 

30' 

.1822 

.2606 

.9833 

.9927 

.1853 

.2680 

5.396 

.7320 

30' 

40' 

.1851 

.2674 

.9827 

.9924 

.1883 

.2750 

5.309 

.7250 

20' 

50' 

.1880 

.2740 

.9822 

.9922 

.1914 

.2819 

5.226 

.7181 

10' 

11° 

.1908 

.2806 

.9816 

.9919 

.1944 

.2887 

5.145 

.7113 

79° 

10' 

.1937 

.2870 

.9811 

.9917 

.1974 

.2953 

5.066 

.7047 

50' 

20' 

.1965 

.2934 

.9805 

.9914 

.2004 

.3020 

4.989 

.6980 

40' 

30' 

.1994 

.2997 

.9799 

.9912 

.2035 

.3085 

4.915 

.6915 

30' 

40' 

.2022 

.3058 

.9793 

.9909 

.2065 

.3149 

4.843 

.6851 

20' 

50' 

.2051 

.3119 

.9787 

.9907 

.2095 

.3212 

4.773 

.6788 

10' 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

A  (in-lst 

Cosine. 

Sine. 

Cotangent. 

Tangent. 

Angle. 

For  angles  over  45°  use  right  column  and  take  names  of  functions  at 
bottom  of  page. 


TABLES 


291 


III.     NATURAL   AND   LOGARITHMIC   FUNCTIONS   (Continued) 


Sine. 

Cosine. 

Tangent. 

Cotangent. 

A      1 

Nat. 

Log. 

,  Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

12° 

0.2079 

9.3179 

0.9781 

9.9904 

0.2126 

9.3275 

4.705 

10.6725 

78° 

10' 

.2108 

.3238 

.9775 

.9901 

.2156 

.3336 

4.638 

.6664 

50' 

20' 

.2136 

.3296 

.9769 

.9899 

.2186 

.3397 

4.574 

.6603 

40' 

30' 

.2164 

.3353 

.9763 

.9896 

.2217 

.3458 

4.511 

.6542 

30' 

40' 

.2193 

.3410 

.9757 

.9893 

.2247 

.3517 

4.449 

.6483 

20' 

50' 

.2221 

.3466 

.9750 

.9890 

.2278 

.3576 

4.390 

.6424 

10' 

13° 

.2250 

.3521 

.9744 

.9887 

.2309 

.3634 

4.332 

.6366 

77° 

10' 

.2278 

.3575 

.9737 

.9884 

.2339 

.3691 

4.275 

.6309 

50' 

20'~ 

.2306 

.3629 

.9730 

.9881 

.2370 

.3748 

4.219 

.6252 

40' 

30' 

.2334 

.3682 

.9724 

.9878 

.2401 

.3804 

4.165 

.6196 

30' 

40' 

.2363 

.3734 

.9717 

.9875 

.2432 

.3859 

4.113 

.6141 

20' 

50' 

.2391 

.3786 

.9710 

.9872 

.2462 

.3914 

4.061 

.6086 

10' 

14° 

.2419 

.3837 

.9703 

.9869 

.2493 

.3968 

4.011 

.6032 

76° 

10' 

.2447 

.3887 

.9696 

.9866 

.2524 

.4021 

3.962 

.5979 

50' 

1  20' 

.2476 

.3937 

.9689 

.9863 

.2555 

.4074 

3.914 

.5926 

40'  §• 

•§  30' 

.2504 

.3986 

.9681 

.9859 

.2586 

.4127 

3.867 

.5873 

30'  -3 

1  40' 

.2532 

.4035 

.9674 

.9856 

.2617 

.4178 

3.821 

.5822 

20'  £ 

£  50' 

.2560 

.4083 

.9667 

.9853 

.2648 

.4230 

3.776 

.5770 

10' 

15° 

.2588 

.4130 

.9659 

.9849 

.2679 

.4281 

3.732 

.5719 

75° 

10' 

.2616 

.4177 

.9652 

.9846 

.2711 

.4331 

3.689 

.5669 

50' 

20' 

.2644 

.4223 

.9644 

.9843 

.2742 

.4381 

3.647 

.5619 

40' 

30' 

.2672 

.4269 

.9636 

.9839 

.2773 

.4430 

3.606 

.5570 

30' 

40' 

.2700 

.4314 

.9628 

.9836 

.2805 

.4479 

3.566 

.5521 

20' 

50' 

.2728 

.4359 

.9621 

.9832 

.2836 

.4527 

3.526 

.5473 

10' 

16° 

.2756 

.4403 

.9613 

.9828 

.2867 

.4575 

3.487 

.5425 

74° 

10' 

.2784 

.4447 

.9605 

.9825 

.2899 

.4622 

3.450 

.5378 

50' 

20' 

.2812 

.4491 

.9596 

.9821 

.2931 

.4669 

3.412 

.5331 

40' 

30' 

.2840 

.4533 

.9588 

.9817 

.2962 

.4716 

3.376 

.5284 

30' 

40' 

.2868 

.4576 

.9580 

.9814 

.2994 

.4762 

3.340 

.5238 

20' 

50' 

.2896 

.4618 

.9572 

.9810 

.3026 

.4808 

3.305 

.5192 

10' 

17° 

.2924 

.4659 

.9563 

.9806 

.3057 

.4853 

3.271 

.5147 

73° 

10' 

.2952 

.4700 

.9555 

.9802 

.3089 

.4898 

3.237 

.5102 

50' 

20' 

.2979 

.4741 

.9546 

.9798 

.3121 

.4943 

3.204 

.5057 

40' 

30' 

.3007 

.4781 

.9537 

.9794 

.3153 

.4987 

3.172 

.5013 

30' 

40' 

.3035 

.4821 

.9528 

.9790 

.3185 

.5031 

3.140 

.4969 

20' 

50' 

.3062 

.4861 

.9520 

.9786 

.3217 

.5075 

3.108 

.4925 

10' 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Cosine. 

Sine. 

Cotangent. 

Tangent. 

Angle. 

For  angles  over  45°  use  right  column  and  take  names  of  functions  at 
bottom  of  page. 


292 


TABLES 


III.     NATURAL   AND   LOGARITHMIC   FUNCTIONS    (Continued) 


Sine. 

Cosine. 

Tangent. 

Cotangent. 

Angle. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

18° 

0.3090 

9.4900 

0.9511 

9.9782 

0.3249 

9.5118 

3.078 

10.4882 

72° 

10' 

.3118 

.4939 

.9502 

.9778 

.3281 

.5161 

3.048 

.4839 

50' 

20' 

.3145 

.4977 

.9492 

.9774 

.3314 

.5203 

3.018 

.4797 

40' 

30' 

.3173 

.5015 

.9483 

.9770 

.3346 

.5245 

2.989 

.4755 

30' 

40' 

.3201 

.5052 

.9474 

.9765 

.3378 

.5287 

2.960 

.4713 

20' 

50' 

.3228 

.5090 

.9465 

.9761 

.3411 

.5329 

2.932 

.4671 

10' 

19° 

.3256 

.5126 

.9455 

.9757 

.3443 

.537C 

2.904 

.4630 

71° 

10' 

.3283 

.5163 

.9446 

.9752 

.3476 

.5411 

2.877 

.4589 

50' 

20' 

.3311 

.5199 

.9436 

.9748 

.3508 

.5451 

2.850 

.4549 

40' 

30' 

.3338 

.5235 

.9426 

.9743 

.3541 

.5491 

2.824 

.4509 

30' 

40' 

.3365 

.5270 

.9417 

.9739 

.3574 

.5531 

2.798 

.4469 

20' 

50' 

.3393 

.5306 

.9407 

.9734 

.3607 

.5571 

2.773 

.4429 

10' 

20° 

.3420 

.5341 

.9397 

.9730 

.3640 

.5611 

2.748 

.4389 

70° 

10' 

.3448 

.5375 

.9387 

.9725 

.3673 

.5650 

2.723 

.4350 

50' 

I  20' 

.3475 

.5409 

.9377 

.9721 

.3706 

.5689 

2.699 

.4311 

40'  * 

I  30' 

.3502 

.5443 

.9367 

.9716 

.3739 

.5727 

2.675 

.4273 

30'  -g 

•§  40' 

.3529 

.5477 

.9356 

.9711 

.3772 

.5766 

2.651 

.4234 

20'  £ 

£  50' 

.3557 

.5510 

.9346 

.9706 

.3805 

.5804 

2.628 

.4196 

10' 

21° 

.3584 

.5543 

.9336 

.9702 

.3839 

.5842 

2.605 

.4158 

69° 

10' 

.3611 

.5576 

.9325 

.9697 

.3872 

.5879 

2.583 

.4121 

50' 

20' 

.3638 

.5609 

.9315 

.9692 

.3906 

.5917 

2.561 

.4083 

40' 

30' 

.3665 

.5641 

.9304 

.9687 

.3939 

.5954 

2.539 

.4046 

30' 

40' 

.3692 

.5673 

.9293 

.9682 

.3973 

.5991 

2.517 

.4009 

20' 

50' 

.3719 

.5704 

.9283 

.9677 

.4006 

.6028 

2.496 

.3972 

10' 

22 

.3746 

.5736 

.9272 

.9672 

.4040 

.6064 

2.475 

.3936 

68° 

10' 

.3773 

.5767 

.9261 

.9667 

.4074 

.6100 

2.455 

.3900 

50' 

20' 

.3800 

.5798 

.9250 

.9661 

.4108 

.6136 

2.434 

.3864 

40' 

30' 

.3827 

.5828 

.9239 

.9656 

.4142 

.6172 

2.414 

.3828 

30' 

40' 

.3854 

.5859 

.9228 

.9651 

.4176 

.6208 

2.395 

.3792 

20' 

50' 

.3881 

.5889 

.9216 

.9646 

.4210 

.6243 

2.375 

.3757 

10' 

23° 

.3907 

.5919 

.9205 

.9640 

.4245 

.6279 

2.356 

.3721 

67° 

10' 

.3934 

.5948 

.9194 

.9635 

.4279 

.6314 

2.337 

.3686 

50' 

20' 

.3961 

.5978 

.9182 

.9629 

.4314 

.6348 

2.318 

.3652 

40' 

30' 

.3987 

.6007 

.9171 

.9624 

.4348 

.6383 

2.300 

.3617 

30' 

40' 

.4014 

.6036 

.9159 

.9618 

.4383 

.6417 

2.282 

.3583 

20' 

50' 

.4041 

.6065 

.9147 

.9613 

.4417 

.6452 

2.264 

.3548 

10' 

Nat. 

Log. 

Nat.  1 

Log. 

Nat. 

Log. 

Nat. 

Log 

A  t,,T|o 

Consie. 

Sine. 

Cotangent. 

Tangent. 

/\IlgH?. 

For  angles  over  45°  use  right  column  and  take  names  of  function  at 
bottom  of  page. 


TABLES 


293 


III.    NATURAL  AND  LOGARITHMIC  FUNCTIONS    (Continued) 


Sine. 

Cosine. 

Tangent. 

Cotangent- 

A  n  LC  1  c  . 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

24° 

0.4067 

9.6093 

0.9135 

9.9607 

0.4452 

9.6486 

2.246 

10.3514 

66° 

10' 

.5094 

.6121 

.9124 

.9602 

.4487 

.6520 

2.229 

.3480 

50' 

20' 

.4120 

.6149 

.9112 

.9596 

.4522 

.6553 

2.211 

.3447 

40' 

30' 

.4147 

.6177 

.9100 

.9590 

.4557 

.6587 

2.194 

.3413 

30' 

40' 

.4173 

.6205 

.9088 

.9584 

.4592 

.6620 

2.178 

.3380 

20' 

50' 

.4200 

.6232 

.9075 

.9579 

.4628 

.6654 

2.161 

.3346 

Iff 

25° 

.4226 

.6259 

.9063 

.9573 

.4663 

.6687 

2.145 

.3313 

65° 

10' 

.4253 

.6286 

.9051 

.9567 

.4699 

.6720 

2.128 

.3280 

50' 

20' 

.4279 

.6313 

.9038 

.9561 

.4734 

.6752 

2.112 

.3248 

40' 

30' 

.4305 

.6340 

.9026 

.9555 

.4770 

.6785 

2.097 

.3215 

30' 

40' 

.4331 

.6366 

.9013 

.9549 

.4806 

.6817 

2.081 

.3183 

20' 

50' 

.4358 

.6392 

.9001 

.9543 

.4841 

.6850 

2.066 

.3150 

10' 

26° 

.4384 

.6418 

.8988 

.9537 

.4877 

.6882 

2.050 

.3118 

64° 

10' 

.4410 

.6444 

.8975 

.9530 

.4913 

.6914 

2.035 

.3086 

50' 

I  20' 

.4436 

.6470 

.8962 

.9524 

.4950 

.6946 

2.020 

.3054 

40'  R 

1  30' 

.4462 

.6495 

.8949 

.9518 

.4986 

.6977 

2.006 

.3023 

30'  , 

1  40' 

.4488 

.6521 

.8936 

.9512 

.5022 

.7009 

1.991 

.2991 

20'  1 

£  50' 

.4514 

.6546 

.8923 

.9505 

.5059 

.7040 

1.977 

.2960 

10'  « 

27° 

.4540 

.6570 

.8910 

.9499 

.5095 

.7072 

1.963 

.2928 

63° 

10' 

.4566 

.6595 

.8897 

.9492 

.5132 

.7103 

1.949 

.2897 

50' 

20' 

.4592 

.6620 

.8884 

.9486 

.5169 

.7134 

1.935 

.2866 

40' 

30' 

.4617 

.6644 

.8870 

.9479 

.5206 

.7165 

1.921 

.2835 

30' 

40' 

.4643 

.6668 

.8857 

.9473 

.5243 

.7196 

1.907 

.2804 

20' 

50' 

.4669 

.6692 

.8843 

.9466 

.5280 

.7226 

1.894 

.2774 

10' 

28° 

.4695 

.6716 

.8829 

.9459 

.5317 

.7257 

1.881 

.2743 

62° 

10' 

.4720 

.6740 

.8816 

.9453 

.5354 

.7287 

1.868 

.2713 

50' 

20' 

.4746 

.6763 

.8802 

.9446 

.5392 

.7317 

1.855 

.2683 

40' 

30' 

.4772 

.6787 

.8788 

.9439 

.5430 

.7348 

1.842 

.2652 

30' 

40' 

.4797 

.6810 

.8774 

.9432 

.5467 

.7378 

1.829 

.2622 

20' 

50' 

.4823 

.6833 

.8760 

.9425 

.5505 

.7408 

1.817 

.2592 

10' 

29° 

.4848 

.6856 

.8746 

.9418 

.5543 

.7438 

1.804 

.2562 

61° 

10' 

.4874 

.6878 

.8732 

.9411 

.5581 

.7467 

1.792 

.2533 

50' 

20' 

.4899 

.6901 

.8718 

.9404 

.5619 

.7497 

1.780 

.2503 

40' 

30' 

.4924 

.6923 

.8704 

.9397 

.5658 

.7526 

1.768 

.2474 

30' 

40' 

.4950 

.6946 

.8689 

.9390 

.5696 

.7556 

1.756 

.2444 

20' 

50' 

.4975 

.6968 

.8675 

.9383 

.5735 

.7585 

1.744 

.2415 

10' 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

A  .,»?,-, 

Cosine. 

Sine. 

Cotangent. 

Tangent. 

Angle. 

For  angles  over  45°  use  right  column  and  take  names  of  functions  at 
bottom  of  page. 


294 


TABLES 


III.    NATURAL  AND   LOGARITHMIC  FUNCTIONS    (Continued) 


Sine. 

Cosine. 

Tangent. 

Cotangent. 

Angle. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

30° 

0.5000 

9.6990 

0.8660 

9.9375 

0.5774 

9.7614 

1.732 

10.2386 

60° 

10' 

.5025 

.7012 

.8646 

.9368 

.5812 

.7644 

1.721 

.2356 

50' 

20' 

.5050 

.7033 

.8631 

.9361 

.5851 

.7673 

1.709 

.2327 

40' 

30' 

.5075 

.7055 

.8616 

.9353 

.5890 

.7701 

1.698 

.2299 

30' 

40' 

.5100 

.7076 

.8601 

.9346 

.5930 

.7730 

1.686 

.2270 

20' 

50' 

.5125 

.7097 

.8587 

.9338 

.5969 

.7759 

1.675 

.2241 

10' 

31° 

.5150 

.7118 

.8572 

.9331 

.6009 

.7788 

1.664 

.2212 

59° 

10' 

.5175 

.7139 

.8557 

.9323 

.6048 

.7816 

1.653 

.2184 

50' 

20' 

.5200 

.7160 

.8542 

.9315 

.6088 

.7845 

1.643 

.2155 

407 

30' 

.5225 

.7181 

.8526 

.9308 

.6128 

.7873 

1.632 

.2127 

30' 

40' 

.5250 

.7201 

.8511 

.9300 

.6168 

.7902 

1.621 

.2098 

20' 

50' 

.5275 

.7222 

.8496 

.9292 

.6208 

.7930 

1.611 

.2070 

10' 

32 

.5299 

.7242 

.8480 

.9284 

.6249 

.7958 

1.600 

.2042 

58° 

10' 

.5324 

.7262 

.8465 

.9276 

.6289 

.7986 

1.590 

.2014 

50' 

§  20' 

.5348 

.7282 

.8450 

.9268 

.6330 

.8014 

1.580 

.1986 

40'  a 

1  30' 

.5373 

.7302 

.8434 

.9260 

.6371 

.8042 

1.570 

.1958 

30'  ^ 

•3  40' 

.5398 

.7322 

.8418 

.9252 

.6412 

.8070 

1.560 

.1930 

20'  § 

£  50' 

.5422 

.7342 

.8403 

.9244 

.6453 

.8097 

1.550 

.1903 

10' 

33° 

.5446 

.7361 

.8387 

.9236 

.6494 

.8125 

1.540 

.1875 

57° 

10' 

.5471 

.7380 

.8371 

.9228 

.6536 

.8153 

1.530 

.1847 

50' 

20' 

.5495 

.7400 

.8355 

.9219 

.6577 

.8180 

1.520 

.1820 

40' 

30' 

.5519 

.7419 

.8339 

.9211 

.6619 

.8208 

1.511 

.1792 

30' 

40' 

.5544 

.7438 

.8323 

.9203 

.6661 

.8235 

1.501 

.1765 

20' 

50' 

.5568 

.7457 

.8307 

.9194 

.6703 

.8263 

1.492 

.1737 

10' 

34° 

.5592 

.7476 

.8290 

.9186 

.6745 

.8290 

1.483 

.1710 

56° 

10' 

.5616 

.7494 

.8274 

.9177 

.6787 

.8317 

1.473 

.1683 

50' 

20' 

.5640 

.7513 

.8258 

.9169 

.6830 

.8344 

1.464 

.1656 

40' 

30' 

.5664 

.7531 

.8241 

.9160 

.6873 

.8371 

1.455 

.1629 

30' 

40' 

.5688 

.7550 

.8225 

.9151 

.6916 

.8398 

1.446 

.1602 

20' 

50' 

.5712 

.7568 

.8208 

.9142 

.6959 

.8425 

1.437 

.1575 

10' 

35° 

.5736 

.7586 

.8192 

.9134 

.7002 

.8452 

1.428 

.1548 

55° 

10' 

.5760 

.7604 

.8175 

.9125 

.7046 

.8479 

1.419 

.1521 

50' 

20' 

.5783 

.7622 

.8158 

.9116 

.7089 

.8506 

1.411 

.1494 

40' 

30' 

.5807 

.7640 

.8141 

.9107 

.7133 

.8533 

1.402 

.1467 

30' 

40' 

.5831 

.7657 

.8124 

.9098 

.7177 

.8559 

1.393 

.1441 

20' 

50' 

.5854 

.7675 

.8107 

.9089 

.7221 

.8586 

1.385 

.1414 

10' 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

A  ,1,1,1 

Cosine. 

Sine. 

Cotangent. 

Tangent. 

Angle. 

For  angles  over  45°  use  right  column  and  take  names  of  functions  at 
bottom  of  page. 


TABLES 


295 


III.    NATURAL  AND   LOGARITHMIC  FUNCTIONS    (Continued) 


Sine. 

Cosine. 

Tangent. 

Cotangent. 

Angle. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

36° 

0.5878 

9.7692 

0.8090 

9.9080 

0.7265 

9.8613 

1.376 

10.1387 

54° 

10' 

.5901 

.7710 

.8073 

.9070 

.7310 

.8639 

1.368 

.1361 

50' 

20' 

.5925 

.7727 

.8056 

.9061 

.7355 

.8666 

1.360 

.1334 

40' 

30' 

.5948 

.7744 

.8039 

.9052 

.7400 

.8692 

1.351 

.1308 

30' 

40' 

.5972 

.7761 

.8021 

.9042 

.7445 

.8718 

1.343 

.1282 

20' 

50' 

.5995 

.7778 

.8004 

.9033 

.7490 

.8745 

1.335 

.1255 

10' 

37° 

.6018 

.7795 

.7986 

.9023 

.7536 

.8771 

1.327 

.1229 

53° 

10' 

.6041 

.7811 

.7969 

.9014 

.7581 

.8797 

1.319 

.1203 

50' 

20' 

.6065 

.7828 

.7951 

.9004 

.7627 

.8824 

1.311 

.1176 

40' 

30' 

.6088 

.7844 

.7934 

.8995 

.7673 

.8850 

1.303 

.1150 

30' 

40' 

.6111 

.7861 

.7916 

.8985 

.7720 

.8876 

1.295 

.1124 

20' 

50' 

.6134 

.7877 

.7898 

.8975 

.7766 

.8902 

1.288 

.1098 

10' 

38° 

.6157 

.7893 

.7880 

.8965 

.7813 

.8928 

1.280 

.1072 

52° 

10' 

.6180 

.7910 

.7862 

.8955 

.7860 

.8954 

1.272 

.1046 

50' 

1  20' 

.6202 

.7926 

.7844 

.8945 

.7907 

.8980 

1.265 

.1020 

40'  a 

-§  30' 

.6225 

.7941 

.7826 

.8935 

.7954 

.9006 

1.257 

.0994 

30'  § 

1  40' 

.6248 

.7957 

.7808 

.8925 

.8002 

.9032 

1.250 

.0968 

20'  1 

£  50' 

.6271 

.7973 

.7790 

.8915 

.8050 

.9058 

1.242 

.0942 

w* 

39° 

.6293 

.7989 

.7771 

.8905 

.8098 

.9084 

1.235 

.0916 

51° 

10' 

.6316 

.8004 

.7753 

.8895 

.8146 

.9110 

1.228 

.0890 

50' 

20' 

.6338 

.8020 

.7735 

.8884 

.8195 

.9135 

1.220 

.0865 

40' 

30' 

.6361 

.8035 

.7716 

.8874 

.8243 

.9161 

1.213 

.0839 

30' 

40' 

.6383 

.8050 

.7698 

.8864 

.8292 

.9187 

1.206 

.0813 

20' 

50' 

.  .6406 

.8066 

.7679 

.8853 

.8342 

.9212 

1.199 

.0788 

10' 

40° 

.6428 

.8081 

.7660 

.8843 

.8391 

.9238 

1.192 

.0762 

50° 

10' 

.6450 

.8096 

.7642 

.8832 

.8441 

.9264 

1.185 

.0736 

50' 

20' 

.6472 

.8111 

.7623 

.8821 

.8491 

.9289 

1.178 

.0711 

40' 

30' 

.6494 

.8125 

.7604 

.8810 

.8541 

.9315 

1.171 

.0685 

30' 

40' 

.6517 

.8140 

.7585 

.8800 

.8591 

.9341 

1.164 

.0659 

20' 

50' 

.6539 

.8155 

.7566 

.8789 

.8642 

.9366 

1.157 

.0634 

10' 

41° 

.6561 

.8169 

.7547 

.8778 

.8693 

.9392 

1.150 

.0608 

49° 

10' 

.6583 

.8184 

.7528 

.8767 

.8744 

.9417 

1.144 

.0583 

50' 

20' 

.6604 

.8198 

.7509 

.8756 

.8796 

.9443 

1.137 

.0557 

40' 

30' 

.6626 

.8213 

.7490 

.8745 

.8847 

.9468 

1.130 

.0532 

30' 

40' 

.6648 

.8227 

.7470 

.8733 

.8899 

.9494 

1.124 

.0506 

20' 

50' 

.6670 

.8241 

.7451 

.8722 

.8952 

.9519 

1.117 

.0481 

10' 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Cosine. 

Sine. 

Cotangent. 

Tangent. 

Angle. 

For  angles  over  45°  use  right  column  and  take  names  of  functions  at 
bottom  of  page. 


296 


TABLES 


III.     NATURAL  AND  LOGARITHMIC  FUNCTIONS     (Continued) 


Sine. 

Cosine. 

Tangent. 

Cotangent. 

Angle. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

42° 

0.6691 

9.8255 

0.7431 

9.8711 

0.9004 

9.9544 

1.111 

10.0456 

48° 

10' 

.6713 

.8269 

.7412 

.8699 

.9057 

.9570 

1.104 

.0430 

50' 

20' 

.6734 

.8283 

.7392 

.8688 

.9110 

.9595 

1.098 

.0405 

40' 

30' 

.6756 

.8297 

.7373 

.8676 

.9163 

.9621 

1.091 

.0379 

30' 

40' 

.6777 

.8311 

.7353 

.8665 

.9217 

.9646 

1.085 

.0354 

20' 

50' 

.6799 

.8324 

.7333 

.8653 

.9271 

.9671 

1.079 

.0329 

10' 

43° 

.6820 

.8338 

.7314 

.8641 

.9325 

.9697 

1.072 

.0303 

47° 

10' 

.6841 

.8351 

.7294 

.8629 

.9380 

.9722 

1.066 

.0278 

50' 

I    20' 

.6862 

.8365 

.7274 

.8618 

.9435 

.9747 

1.060 

.0253 

40'   ft 

1    30' 

.6884 

.8378 

.7254 

.8606 

.9490 

.9772 

1.054 

.0228 

30'   " 

1    40' 

.6905 

.8391 

.7234 

.8594 

.9545 

.9798 

1.048 

.0202 

20'  1 

3    50' 

.6926 

.8405 

.7214 

.8582 

.9601 

.9823 

1.042 

.0177 

10'* 

44° 

.6947 

.8418 

.7193 

.8569 

.9657 

.9848 

1.036 

.0152 

46° 

10' 

.6967 

.8431 

.7173 

.8557 

.9713 

.9874 

1.030 

.0126 

50' 

20' 

.6988 

.8444 

.7153 

.8545 

.9770 

.9899 

1.024 

.0101 

40' 

30' 

.7009 

.8457 

.7133 

.8532 

.9827 

.9924 

1.018 

.0076 

30' 

40' 

.7030 

.8469 

.7112 

.8520 

.9884 

.9949 

1.012 

.0051 

20' 

50' 

.7050 

.8482 

.7092 

.8507 

.9942 

.9975 

1.006 

.0025 

10' 

45° 

.7071 

.8495 

.7071 

.8495 

1.0000 

.0000 

1.0JDO 

.0000 

45° 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

A    _  _,!_ 

Cosine. 

Sine. 

Cotangent. 

Tangent. 

Angle. 

For  angles  over  45°  use  right  column  and  take  names  of  functions  at 
bottom  of  page. 


TABLES 


297 


IV.    NAPERIAN  LOGARITHMS  OF  NUMBERS 
1  to  9.9 


No. 

0 

1 

1 

3 

4 

5 

6 

7 

8 

9 

1 

0.0000 

0.0953 

0.0182 

0.2624 

0.3365 

0.4055 

0.4700 

0.5306 

0.5878 

0.6419 

2 

0.6931 

0.7419 

0.7885 

0.8329 

0.8755 

0.9163 

0.9555 

0.9933 

1.0296 

1.0647 

3 

1.0986 

1.1314 

1  .  1632 

1.1939 

1.2238 

1.2528 

1.2809 

1.3083 

1.3350 

1.3610 

4 

1.3863 

1.4110 

1.4351 

1.4586 

1.4816 

1.5041 

1.5261 

1.5476 

1.5686 

1.5892 

5 

1.6094 

1.6292 

1.6487 

1.6677 

1.6864 

1.7047 

1.7228 

1.7405 

1.7579 

1.7750 

6 

1.7918 

1.8083 

1.8245 

1.8406 

1.8563 

1.8718 

1.8871 

1.9021 

1.9169 

1.9315 

7 

1.9459 

1.9601 

1.9741 

1.9879 

2.0015 

2.0149 

2.0281 

2.0412 

2.0541 

2.0669 

8 

2.0794 

2.0906 

2.1041 

2.1163 

2.1282 

2.1401 

2.1518 

2.1633 

2.1748 

2.1861 

9 

2.1972 

2.2083 

2.2192 

2.2300 

2.2407 

2.2513 

2.2618 

2.2721 

2.2824 

2.2925 

1 

2.3026 

2.3979 

2.4849 

2.5649 

2.6391 

2.7081 

2.7726 

2.8332 

2.8904 

2.9444 

2 

2.9957 

3.0445 

3.0910 

3.1355 

3.1781 

3.2189 

3.2581 

3.2958 

3.3322 

3.3673 

3 

3.4012 

3.4340 

3.4657 

3.4965 

3.5264 

3.5553 

3.5835 

3.6109 

3.6376 

3.6635 

4 

3.6889 

3.7136 

3.7377 

3.7602 

3.7842 

3.8067 

3.8286 

3.8501 

3.8712 

3.8918 

5 

3.9120 

3.9318 

3.9512 

3.9703 

3.9890 

4.0073 

4.0254 

4.0431 

4.0604 

4.0775 

6 

4.0943 

4.1109 

4.1271 

4.1431 

4.1589 

4.1744 

4.1897 

4.2047 

4.2195 

4.2341 

7 

4.2485 

4.2627 

4.2767 

4.2905 

4.3041 

4.3175 

4.3307 

4.3488 

4.3567 

4.3694 

8 

4.3820 

4.3944 

4.4067 

4.4188 

4.4308 

4.4427 

4.4543 

4.4659 

4.4773 

4.4886 

9 

4.4998 

4.5109 

4.5218 

4.5326 

4.5433 

4.5539 

4.5643 

4.5747 

4.5850 

4.5951 

V.    CONVERSION  TABLES 


Radians  to  degrees. 

Degrees  to  radians. 

Grades  to  degrees. 

Mils  to  degrees. 

0.1 

5°  44' 

0°  10' 

0.00291 

0.1 

0°    5.4' 

1 

0°    3'  22.  5" 

0.2 

11    28 

0    20 

0.00582 

0.2 

0    10.8 

2 

0     6  45.0 

0.3 

17    11 

0    30 

0.00873 

0.3 

0    16.2 

3 

0    10     7.5 

0.4 

22    55 

1 

0.01745 

0.4 

0   21.6 

4 

0    13  30.0 

0.5 

28   39 

2 

0.03491 

0.5 

0   27.0 

5 

0    16  52.5 

0.6 

34   23 

3 

0.05236 

0.6 

0   32.4 

6 

0   20  15.0 

0.7 

40    06 

4 

0.06981 

0.7 

0   37.8 

7 

0   23  37.5 

0  8 

45   50 

5 

0.08727 

0.8 

0   43.2 

8 

0   27  00.0 

0.9 

51    34 

10 

0.17453 

0.9 

0   48.6 

9 

0   30  22.5 

1.0 

57    18 

20 

0.34907 

1.0 

0   54.0 

10 

0   33  45.0 

2.0 

114   35 

30 

0.52360 

2.0 

1    48.0 

15 

0   50  37.5 

3.0 

171    53 

40 

0.69813 

3.0 

2   42.0 

20 

1     7  30.0 

4.0 

229    11 

50 

0.87266 

4.0 

3   36.0 

25 

1    24  22.5 

5.0 

286   29 

57    18 

1.00000 

5.0 

4   30.0 

30 

1    41   15.0 

6.0 

343    46 

60 

1.04720 

6.0 

5   24.0 

35 

1    58     7.5 

7.0 

401    04 

90 

1.57080 

7.0 

6    18.0 

40 

2    15  00.0 

8.0 

458   22 

8.0 

7    12.0 

50 

2   48  45.0 

9,0 

515    40 

9.0 

8   06.0 

60 

3   22  30.0 

10.0 

9   00.0 

70 

3    56  15.0 

20.0 

18   00.0 

80 

4   30  00.0 

30.0 

27    00.0 

90 

5     3  45.0 

40.0 

36   00.0 

50  0 

45    00  0 

100.0 

90   00.0 

298 


TABLES 


VI.    FUNCTIONS  OF  ANGLES   IN  GRADES    

Since  100  grades  equals  a  quadrant  or  90°,  functions  of  angles  greater  than  100  grades  can  be 
found  from  this  table  by  use  of  formulas  in  Chapter  VIII  for  finding  functions  of  all  angles 
in  terms  of  functions  of  angles  less  than  90°. 


Sine. 

Cosine. 

Tangent. 

Cotangent. 

Angle, 

grades. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

0 

0.0000 

—  00 

1.000 

10.0000 

0.0000 

—  00 

00 

00 

100 

1 

.0157 

8.1961 

0.9997 

9.9999 

.01571 

8.1962 

63.66 

11.8058 

99 

2 

.0314 

.4971 

.9995 

.9998 

.0314 

.4973 

31.82 

.5027 

98 

3 

.0471 

.6731 

.9989 

.9995 

.0472 

.6736 

21.20 

.3264 

97 

4 

.0628 

.7979 

.9979 

.9991 

.0629  • 

.7988 

15.90 

.2013 

96 

5 

.0786 

.8946 

.9970 

.9987 

.0787 

.8960 

12.71 

.1040 

95 

6 

.0941 

.9736 

.9955 

.9981 

.0945 

.9756 

10.58 

.0244 

94 

7 

.1097 

9.0403 

.9940 

.9974 

.1104 

9.0430 

9.057 

10.9570 

93 

8 

.1254 

.0981 

.9922 

.9966 

.1263 

.1015 

7.916 

.8985 

92 

9 

.1409 

.1489 

.9901 

.9957 

.1423 

.1533 

7.026 

.8467 

91 

10 

.1564 

.1943 

.9876 

.9946 

.1584 

.1997 

6.314 

.8003 

90 

11 

.1719 

.2354 

.9851 

.9935 

.1745 

.2419 

5.729 

.7581 

89 

12 

.1874 

.2727 

.9822 

.9922 

.1908 

.2805 

5.242 

.7195 

88 

13 

.2028 

.3070 

.9774 

.9909 

.2071 

.3162 

4.828 

.6838 

87 

14 

.2181 

.3387 

.9759 

.9894 

.2235 

.3493 

4.474 

.6507 

86 

15 

.2335 

.3682 

.9723 

.9878 

.2401 

.3804 

4.166 

.6197 

85 

16 

.2487 

.3957 

.9685 

.9861 

.2567 

.4095 

3.895 

.5905 

84 

17 

.2639 

.4214 

.9645 

.9843 

.2736 

.4370 

3.655 

.5629 

83 

18 

.2790 

.4456 

.9603 

.9824 

.2905 

.4632 

3.442 

.5368 

82 

19 

.2940 

.4684 

.9559 

.9804 

.3076 

.4880 

3.251 

.5120 

81 

20 

.3091 

.4900 

.9510 

.9782 

.3249 

.5118 

3.077 

.4882 

80 

21 

.3239 

.5104 

.9460 

.9759 

.3424 

.5345 

2.921 

.4655 

79 

B  22 

.3388 

.5299 

.9408 

.9735 

.3600 

.5563 

2.778 

.4438 

78 

I  23 

.3535 

.5484 

.9354 

.9710 

.3778 

.5773 

2.647 

.4227 

77 

4  24 

.3681 

.5660 

.9298 

.9684 

.3959 

.5976 

2.526 

.4024 

76 

•g  25 

.3827 

.5828 

.9239 

.9656 

.4142 

75172 

2.414 

.3828 

75  §• 

1  26 

.3972 

.5990 

.9177 

.9627 

.4327 

.6362 

2.311 

.3638 

74  -s 

05  27 

.4115 

.6144 

.9114 

.9597 

.4515 

.6547 

2.215 

.3453 

73  S 

28 

.4258 

.6292 

.9049 

.9566 

.4705 

.6726 

2.125 

.3274 

72  « 

29 

.4400 

.6434 

.8981 

.9533 

.4899 

.6901 

2.042 

.3100 

71 

30 

.4540 

.6571 

.8910 

.9499 

.5096 

.7072 

1.962 

.2928 

70 

31 

.4680 

.6702 

.8837 

.9463 

.5294 

.7238 

1.889 

.2762 

69 

32 

.4817 

.6828 

.8764 

.9427 

.5498 

.7402 

1.819 

.2598 

68 

33 

.4955 

.6950 

.8686 

.9388 

.5704 

.7562 

1.753 

.2438 

67 

34 

.5091 

.7068 

.8608 

.9349 

.5914 

.7719 

1.690 

.2281 

66 

35 

.5225 

.7181 

.8527 

.9308 

.6128 

.7873 

1.632 

.2127 

65 

36 

.5358 

.7290 

.8443 

.9265 

.6346 

.8TJ25 

1.576 

.1975 

64 

37 

.5490 

.7396 

.8358 

.9221 

.6569 

.8175 

1.522 

.1825 

63 

38 

.5621 

.7498 

.8272 

.9176 

.6797 

.8323 

1.472 

.1678 

62 

39 

.5750 

.7597 

.8181 

.9128 

.7028 

.8468 

1.423 

.1532 

61 

40 

.5878 

.7692 

.8091 

.9080 

.7266 

.8613 

1.376 

.1387 

60 

41 

.6005 

.7785 

.7997 

.9029 

.7508 

.8755 

1.332 

.1245 

59 

42 

.6129 

.7874 

.7901 

.8977 

.7757 

.8897 

1.289 

.1103 

58 

43 

.6253 

.7961 

.7804 

.8923 

.7950 

.9037 

1.248 

.0963 

57 

44 

.6374 

.8044 

.7706 

.8868 

.8274 

.9177 

1.209 

.0824 

56 

45 

.6494 

.8125 

.7605 

.8811 

.8541 

.9315 

1.171 

.0685 

55 

46 

.6613 

.8204 

.7501 

.8751 

.8817 

.9453 

1.134 

.0547 

54 

47 

.6730 

.8280 

.7396 

.8690 

.9099 

.9590 

1.099 

.0410 

53 

48 

.6845 

.8354 

.7290 

.8627 

.9391 

.9727 

1.065 

.0273 

52 

49 

.6960 

.8426 

.7181 

.8562 

.9692 

.9864 

1.032 

.0137 

51 

50 

.7071 

.8495 

.7071 

.8495 

1.0000 

10.0000 

1.000 

10.0000 

50 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Nat. 

Log. 

Angle, 

Cosine. 

Sine. 

Cotangent. 

Tangent. 

grades. 

For  angles  over  50  grades  read  angles  on  right  and  take  names  of  columns  at  bottom  of 


page. 


TABLES 


299 


VII.    NATURAL  SINES  AND  COSINES  —  ANGLES 
EXPRESSED  IN  MILS 


Mila. 

Degrees. 

Minutes. 

Sine. 

Cosine. 

Mils. 

0 

0 

0 

0 

1.0000 

1600 

50 

2 

48.75 

.0490 

.9988 

1550 

100 

5 

37.50 

.0980 

.9952 

1500 

150 

8 

26.25 

.1467 

.9892 

1450 

200 

11 

15.00 

.1951 

.9808 

1400 

250 

14 

03.75 

.2430 

.9700 

1350 

300 

16 

52.50 

.2903 

.9569 

1300 

350 

19 

41.25 

.3369 

.9415 

1250 

400 

22 

30.00 

.3827 

.9239 

1200 

450 

25 

18.75 

.4273 

.9040 

1150 

500 

28 

07.50 

.4714 

.8819 

1100 

550 

30 

56.25 

.5141 

.8577 

1050 

600 

33 

45.00 

.5556 

•  .8315 

1000 

650 

36 

33.75 

.5957 

.8032 

950 

700 

39 

22.50 

.6344 

.7731 

900 

750 

42 

11.25 

.6716 

.7410 

850 

800 

45 

00.00 

.7071 

.7071 

800 

Mils. 

Degrees. 

Minutes. 

Sine. 

Cosine. 

Mils. 

*. 


INDEX 


(Numbers  refer  to  pages) 


Abscissa,  37 

Absolute  value,  66,  122 

Acceleration,  definition  of,  215 

derivative  of  velocity,  215 
Aggregation,  signs  of,  1 
Amplitude,  measurement  of,  123 
Angle,  between  two  curves,  215 

eccentric,  185,  186 

measurement  of,  104 

mil,  106 

of  depression,  107 

of  elevation,  107 

radian,  105 

vectorial,  120 
Arc,  length  of,  268 
Arithmetic  progression,  220 
Axes,  coordinate,  37 

major  and  minor,  181 

Binomial  formula,  15 
series,  234 

Calculation,  methods  of,  20 

by  geometric  method,  22 

by  logarithms,  25 

by  interpolation,  31,  39,  42,  43, 
113 

by  slide  rule,  30,  31 
Centroid,  264 

Cologarithm,  definition  of,  26 
Complex  number,  addition  of,  123 

definition  of,  63 

division  of,  124 

graphic  representation,  122 


Complex  number,  modulus  of,  122 

multiplication  of,  123 

subtraction  of,  123 
Conic,  definition  of,  177 

center  of,  189 

confocal,  188 

diameter  of,  183 

graphs  of,  164,  165,  166,  178,  179 
Coordinates,  definition  of,  38 

abscissa,  37 

number  pairs,  53,  69,  70 

ordinate,  38 

polar,  120 
Cosine  law,  95 

example,  97 
Critical  value,  207 

determination  of,  207 
Cube  root,  table  of,  284 
Curve,  area  under,  256 

slope  of,  202 

angle  between  two,  204 

Departure,  definition  of,  263 
Depression,  angle  of,  107 
Derivative,  definition  of,  194 

slope  of  a  curve,  202 

to  define  motion,  215 
Diameter  of  a  curve,  183 

conjugate,  184 

Differential,  definition  of,  263 
Directrix,  177 

Eccentricity,  e,  177 
Elevation,  angle  of,  107 


301 


302 


INDEX 


Ellipse,  definition  of,  179 

eccentric  angle  of,  185 

equation  of,  180 

major  and  minor  axes  of,  180 
Empirical  formulas,  57,  244,  245 
Equations,  equal  roots,  217 

equivalence  of,  169 

exponential,  28 

general     form,     second     degree, 
187 

quadratic,  13,  155 

roots  of,  8,  139,  140,  141 

simultaneous,  9,  13,  169 

solution  of,  8 
Equilibrant,  264 
Equilibrium,  of  particle,  132,  133 

of  rigid  body,  133 

Factoring,  Euclidean  method,  5 

formulas  relating  to,  2,  11 
Force,  components  of,  130 

moment  of,  130,  133 
Function,  average  value  of,  259 

continuous,  65 

definition  of,  65,  69,  70 

derivative  of,  194 

even,  241 

expansion  of,  in  series,  232 

hyperbolic,  168 

increasing,  207 

implicit  and  explicit,  163 

integral,  139 

irrational,  168 

linear,  152 

logarithmic,  237,  244 

maximum  value  of,  205 

minimum  value  of,  205 

odd,  241     * 

rational  fractional,  167 

roots  of,  139 

theorems  on,  140 

trigonometric,  72 

zero  and  infinity  of,  141 


Geometric  progression,  222 
Grade,  definition  of,  table  of,  298 
Graphs,  36 

construction  of,  54 

of  functions,  53 

of  trigonometric  functions,  108 

representation  by,  36,  205 
Gravity,  center  ol,  264 

Haversine,  definition  of,  73,  115 
Highest  Common  Factor,  5 
Hyperbola,  definition  of,  181 

conjugate,  182 

diameter  of,  184 

eccentric  angle  of,  186 

equation  of,  165,  182 

Imaginary  Quantity,  (i),  11,  63 
Increment,  66,  195 
Index  Laws,  2 

negative  exponents,  2 

zero  exponent,  2,  11 
Inequalities,  14 
Infinity,  65 
Infinitesimal,  65 
Integral,  definite,  254 

indefinite,  254 

table  of,  276 
ntegration,  250 

formulas  of,  251 

by  parts,  267 

by  substitution,  269 
Interpolation,  methods  of,  31 

by  means  of  graphs,  39,  42,  43 

double,  31 

simple,  280 

special  forms  of,  113 

Latitude,  definition  of,  114 
Latus  rectum,  definition  of,  178 
Limit,  definition  of,  64 

quadrant,  78 

theorems  on,  66,  193 


INDEX 


303 


Lines,  angle  between,  160 
distance  between,  160 
division  of,  158 
equations  of,  157 

general  form,  152,  157 

normal  form,  157 

one-point-slope  form,  157 

slope-intercept  form,  157 

slope  of,  38,  153 

two-point  form,  158 

theorem  on,  152 
Logarithm,  change  of  base,  237 
characteristic  of,  27 
definition  of,  25 
graph  of  function,  244 
idea  of,  24 

logarithmic  paper,  246 
mantissa,  27 
modulus,  237 
Naperian  base,  297 
of  trigonometric  functions,  288 
rules  for  calculation,  26 
semi-logarithmic  paper,  248 
table  of  numbers,  286 
Lowest  Common  Multiple,  4 

Maximum  value  of  function,  205, 

206,  208,  209 
Mil,  definition  of,  106 

table  of,  299 

Minimum  value  of  function,  205 
Motion,  214 

Number,  63 
complex,  63,  123 
imaginary,  63 
rational  and  irrational,  62 
reciprocal  of,  23,  34 
square  roots  of,  24,  284 
scalar,  128 
vector,  123 

Ordinate,  37 


Parabola,  164,  177 
Parameter,  definition  of,  134 
Particle,  definition  of,  133 

equilibrium  of,  133 
Points,  coordinates  of,  38 

distance  between,  158 
Proportion,  definition  of,  45 

theorems  on,  46 

Quadrant  Limits,  78 

Radian,  definition  of,  45 
Radicals,  formulas  relating  to,  11 
Ratio,  definition,  45 
Reciprocal  of  number,  23,  34 
Remainder  Theorem,  143 
Resultant,  264 

Scalar,  128 

definition  of,  125 

product,  128 
Sequence,  64 

arithmetic,  219 

geometric,  219 
Series,  arithmetic,  220 

binomial,  234 

expansion  of  function  in,  232 

exponential,  236 

geometric,  222 

harmonic,  225 

harmonic  mean,  226 

power,  232 

ratio  test  of,  230 

sum  of  terms  of,  220 

with  complex  terms,  231 
Slide  Rule,  30 

rules  for,  31 
Slope,  definition  of,  38 

by  derivative,  202  - 

of  line,  38,  154 
Sine  Law,  95 

Solid  of  revolution,   definition  of, 
257 

volume  of,  257 


304 


INDEX 


Speed, 214 
Square  root,  24 
table  of,  284 
Synthetic  Division,  142 

Tables,  conversion,  297 

explanation  of,  280 

functions  of  angles  in  grades,  298 

integrals,  276 

logarithms  of  numbers,  286 

Naperian  logarithms,  297 

natural    and    logarithmic    func- 
tions, 289 

natural  sines  and  cosines  in  mils, 
299 

powers  and  roots,  284 
Tangent  Law,  99 

example,  99 

Theorems,  geometrical,  16 
Transformations,  linear,  174 

of  origin,  175 

of  variable,  269 


Transformations,  rotation  of  axes, 

174 

Trigonometric  Functions,  addition 
theorems,  85 

complement  relations,  80 

conversion  formulas,  98 

definitions,  73 

fundamental  formulas,  75 

graphs  of,  108 

inverse,  90 

Variable,  63 

continuous,  63 

dependent  and  independent,  65 
Variation,  46 
Vector,  addition  of,  126 

definition  of,  123 

polygon  of,  127 

product  of,  128 

radius,  120 

Work,  definition  of,  133,  261 
of  variable  force,  261 


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OCT151979 


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